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PLANE  TRIGONOMETRY 


COLLEGES  AND   SECONDARY  SCHOOLS 


BT 


DANIEL  A.  MURRAY,  B.A.,  Ph.D. 
/ »       . 

INSTRUCTOR  IN  MATHEMATICS  IN    CORNELL   UNIVERSITY 
FORMERLY  SCHOLAB  AND  FELLOW  AT  JOHNS  HOPKINS  UNIVERSITY 


or  THf 

VNIYER8ITY 

LONGMANS,    GREEN,    AND    CO. 

91  AND  93  FIFTH  AVENUE,  NEW  YORK 
LONDON  AND  BOMBAY 

190X 


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Copyright,  1899, 
fte  lONGMANS,  GREEN,  AND  OU 

ALL  EIGHTS  EE8ERVED. 


First  Edition,  November,  1899. 

Keprinted,  February  and  September,  1900, 

September,  1901, 


PEEFACE. 


Although  there  are  already  many  excellent  text-books  on 
trigonometry,  there  appears  still  to  be  room  for  one  which  shall 
avoid  the  extremes  of  expansion  and  brevity.  Some  of  the  most 
thorough  and  scholarly  of  these  contain  a  great  variety  of  matters 
which  it  is  impossible  to  consider  in  the  time  usually  assigned  to 
this  study  in  school  and  college.  On  the  other  hand,  the  expla- 
nations given  in  many  other  works  are  so  meagre  that  the  stu- 
dent is  perplexed  and  bewildered  by  the  new  ideas  which  are  so 
abruptly  forced  upon  him,  and  the  difficulties  of  the  teacher  are 
greatly  increased.  The  manner  of  presentation  adopted  in  this 
volume  necessitates  more  reading  matter,  and,  consequently,  a 
somewhat  larger  number  of  pages  than  is  found  in  many  of  the 
recent  text-books  on  trigonometry.  This  has  seemed  unavoid- 
able, however,  for  the  general  consensus  of  opinion  among  those 
with  whom  the  author  has  conferred,  is  that  it  is  essential  to 
explain  in  some  detail  the  principles  of  the  science,  in  order 
that  it  may  be  clearly  and  intelligently  understood  by  an  ele- 
mentary student. 

With  regard  to  the  scope  of  the  book,  it  may  be  said  that  it 
deals  with  the  subjects  considered  in  the  ordinary  course  in  plane 
trigonometry  in  colleges  and  secondary  schools.  It  discusses  the 
topics  usually  required  for  teachers'  certificates,  for  entrance  to 
college,  and  for  examinations  in  trigonometry  in  the  first  year 
of  the  college  curriculum.  It  treats  of  all  the  topics  that  one 
who  has  taken  a  few  months'  course  in  trigonometry  may  be 
reasonably  expected  to  know. 


VI  PBEFACE. 

Careful  consideration  has  been  given  both  to  the  early  diffi- 
culties and  to  the  possible  future  needs  of  the  beginner.  The 
book  differs  somewhat  from  other  text-books  on  this  branch  of 
mathematics  both  in  the  arrangement  and  in  the  manner  of  pre- 
sentation. The  oldest  and  simplest  part  of  trigonometry,  namely, 
the  solution  of  triangles  and  the  associated  practical  problems, 
is  concluded  before  the  more  general  and  abstract  portions  of  the 
study  are  introduced.  The  first  chapters  of  the  book  contain 
little  more  about  trigonometric  ratios  and  angular  analysis  than 
is  sufficient  to  enable  the  beginner  to  understand  clearly  the 
arithmetical  part  of  the  science,  and  its  simple  practical  applica- 
tions. This  arrangement  seems  to  have  several  advantages.  The 
subject  is  rendered  far  less  strange  at  the  beginning,  and,  by 
means  of  practical,  concrete  examples,  the  student  becomes 
familiar  with  the  trigonometric  functions  before  proceeding  to 
the  more  general  treatment.  His  progress  is  thus  made  easier 
and  more  rapid.  Teachers  who  prefer  a  wider  generality  of 
treatment  at  the  outset,  however,  can  select  the  chapters  in  a 
different  order  from  that  followed  in  the  text. 

An  endeavour  has  been  made  to  introduce  the  several  topics  in 
such  a  way  that  the  pupil  may  have,  from  the  very  start,  an  intel- 
ligent idea  of  each  step  in  advance,  as  well  as  of  the  ultimate 
purpose  of  the  study.  In  some  cases,  especially  in  Chapter  II. 
(the  first  chapter  on  trigonometry),  care  has  been  taken  to  pre- 
pare the  mind  of  the  learner  for  the  reception  of  new  ideas,  by 
the  preliminary  solution  of  easy  familiar  exercises.  Throughout 
the  work  the  author  has  endeavoured  to  make  each  step  clear,  and 
thus  to  prevent  the  appearance  of  that  puzzled  feeling  which  has 
such  a  depressing  influence  on  those  entering  upon  a  new  study. 
On  the  other  hand,  he  has  sought  to  develop  independence  of 
mind  and  the  power  of  mental  initiative  on  the  part  of  the  stu- 
dent.   Suggestions  as  to  practical  methods  of  work  are  frequently 


PREFACE.  Vll 

introduced,  and  summaries  are  made  in  several  places  for  the 
purpose  of  helping  the  pupil  to  get  a  better  idea  of  the  subject 
as  a  whole. 

In  the  practical  applications,  marked  attention  has  been  given 
to  the  graphical  method  of  solution,  as  well  as  to  the  method  of 
computation.  The  former  method  serves  as  a  check  upon  the 
latter,  and  affords  practice  in  neat  and  careful  drawing.  What 
is  perhaps  more  important,  however,  is  that  the  students  will  thus 
become  accustomed  to  a  method  which  will  be  used  by  them  in 
other  studies,  and  which  is  often  employed  in  practical  work  by 
engineers  and  others. 

Logarithms  are  used  almost  at  the  beginning  of  the  study  as 
here  presented.  For  this  reason,  and  in  order  to  avoid  making  a 
digression  later  on,  an  introductory  chapter  is  devoted  to  a  review 
on  logarithms.  Examples,  simple  ones  as  a  rule,  are  given  in  the 
several  articles.  Questions  and  exercises  suitable  for  practice  and 
review  on  the  separate  chapters  are  placed  at  the  end  of  the  book 
instead  of  at  the  ends  of  the  chapters.  These  collections  will  be 
found  useful,  both  in  the  short  reviews  that  may  be  required  on 
the  completion  of  each  chapter  and  in  the  larger  and  more  general 
reviews.  Many  of  the  examples  have  been  taken  from  examina- 
tion papers  set  in  Great  Britain  and  the  United  States. 

Throughout  the  work  there  are  many  historical  and  other  notes ; 
and  an  historical  sketch  is  given  in  the  Appendix.  It  is  believed 
that  some  knowledge  of  the  historical  development  of  trigonome- 
try, and  of  the  men  of  various  times  and  races  who  have  helped 
to  advance  the  subject,  will  interest  and  stimulate  those  who  are 
entering  upon  its  study. 

While  writing  this  book,  the  author  has  received  many  valuable 
suggestions  from  Mr.  J.  A.  Clark,  B.S.,  of  the  Ithaca  High  School, 
and  from  several  of  his  colleagues  in  the  departments  of  mathe- 
matics and  of  engineering  at  Cornell  University.     He  is  indebted 


Vlll  PREFACE, 


^ 


to  Dr.  G.  A.  Miller  and  Dr.  J.  V.  Westfall,  of  the  department  of 
mathematics  at  Cornell  University,  for  their  kind  assistance  in! 
the  revision  of  the  proof-sheets,  and  to  Mr.  E.  A.  Miller,  B.S.,  for 
his  friendly  aid  in  working  examples.  The  drawings  have  been 
made  by  Mr.  A.  T.  Bruegel,  M.M.E.,  formerly  instructor  in  the 
kinematics  of  machinery  at  Cornell  University,  now  of  the  Pratt 
Institute,  Brooklyn,  N.Y.  The  author  uses  this  opportunity  to 
express  his  thanks  for  the  pains  taken  by  Mr.  Bruegel  to  make 
the  figures  a  pleasing  feature  of  the  book. 

D.  A.  MUERAY. 

Cornell  University, 
August,  1899. 


CONTENTS. 


CHAPTER  I. 
Review  op  Logarithms. 

ABT.  PAGE 

2.  Definition  of  a  logarithm 2 

3.  Properties  of  logaritiims  .    ~ 2 

4.  Common  system  of  logaritlims 4 

]  5.  Negative  characteristics .5 

6,   Exercises  in  logarithmic  computation .6 


) 


CHAPTER  II. 
Trigonometric  Ratios  of  Acute  Angles. 

8.  Ratio.     Measure .9 

9.  Incommensurable  quantities.     Approximations       ....  12 

10.  Linear  measure.   Drawing  to  scale.  Direct  measurement  by  means 

of  drawing 15 

11.  Degree  measure.     The  protractor 18 

12.  Trigonometric  ratios  defined  for  acute  angles 20 

13.  Definite  and  invariable  connection  between  acute  angles  and  trig- 

onometric ratios 24 

14.  Practical  problems 26 

15.  Trigonometric  ratios  of  45°,  60°,  30°,  0°,  90° 29 

16.  Relations  between  the  trigonometric  ratios  of  an  angle  and  those 

of  its  complement          . 32 

17.  Exponents  in  trigonometry       . 32 

18.  Relations  between  the  trigonometric  ratios  of  an  acute  angle  .        .  33 
9.   Summary 37 

iz 


CONTENTS. 


CHAPTER  III. 

Solution  of  Right-angled  Triangles. 

AKT.  PAGE 

*20.  Solution  of  a  triangle 38 

21.  The  graphical  method 38 

22.  The  method  of  computation 39 

23.  Comparison  between  the  graphical  method  and  the  method  of 

computation 40 

24.  General  directions  for  solving  problems 41 

0-26.    Solution  of  right-angled  triangles .41 

26.  Checks  upon  the  accuracy  of  the  computation         .        .        .        .43 

27.  Cases  in  the  solution  of  right-angled  triangles  .        .        .        .43 


CHAPTER  IV. 
Applications  involving  the  Solution  of  Right-angled  Triangles. 

28.  Projection  of  a  straight  line  upon  another  straight  line   ...  49 

29.  Measurement  of  heights  and  distances 50 

30.  Problems  requiring  a  knowledge  of  the  points  of  the  mariner's 

compass 63 

31.  Mensuration      .         . 64 

32.  Solution  of  isosceles  triangles 55 

33.  Related  regular  polygons  and  circles 55 

34.  Solution  of  oblique  triangles 67 

34a.   Area  of  a  triangle  in  terms  of  its  sides 60 

346.   Distance  and  dip  of  the  visible  horizon  .        .        .        .        .        .01 

34c.   Examples  in  the  measurement  of  land 01 

35.  Summary 63 


CHAPTER   V. 
Trigonometric  Ratios  of  Angles  in  General. 

36.  Directed  lines 65 

37.  Trigonometric  definition  of  an  angle.    Angles  unlimited  in  magni- 

tude.   Positive  and  negative  angles 67 

38.  Supplement  and  complement  of  an  angle 69 

39.  The  convention  of  signs  on  a  plane 70 


40.    General  definition  of  the  trigonometric  ratios 71 


CONTENTS,  xi 


41.  The  algebraic  signs  of  the  trigonometric  ratios  for  angles  in  the 

different  quadrants 72 

42.  To  represent  the  angle  geometrically  when  the  ratios  are  given       .      74 

43.  Connection  between  angles  and  trigonometric  ratios        ...      75 

44.  Relations  between  the  trigonometric  ratios  of  an  angle    .        .        .77 

45.  Ratios  of  90°  -  A,  90°  +  A,  180°  ^  Ay  -  A 79 


CHAPTER  VI. 

Trigonometric  Ratios  of  the  Sum  and  the  Difference  op 
Two  Angles. 

46.  Derivation  of  the  sine  and  cosine  of  the  sum  of  two  angles  when 

each  of  the  angles  is  less  than  a  right  angle  .        .        .        .        .85 

47.  Derivation  of  the  sine  and  cosine  of  the  difference  of"  two  angles 

when  each  of  the  angles  is  less  than  a  right  angle         ...      87 

48.  Proof  of  the  addition  and  subtraction  formulas  for  all  values  of  the 

two  angles 89 

49.  Each  fundamental  formula  contains  the  others        ....  90 

50.  Ratios  of  an  angle  in  terms  of  the  ratios  of  its  half  angle        .        .  90 

51.  Tangents  of  the  sum  and  difference  of  two  angles,  and  of  twice  an 

angle 92 

52.  Sums  and  differences  of  sines  and  cosines 93 


CHAPTER  VII. 
Solution  of  Triangles  in  General. 

53.  Cases  for  solution .        .97 

54.  Fundamental  relations  between  the  sides  and  angles  of  a  triangle. 

The  law  of  sines.    The  law  of  cosines 98 

54a.    Substitution  of  sines  for  sides,  and  of  sides  for  sines     .        .        .  101 

55.  Case  I.     Given  one  side  and  two  angles 101 

56.  Case  II.     Given  two  sides  and  an  angle  opposite  to  one  of  them     .  102 

57.  Case  III.    Given  two  sides  and  their  included  angle         .        .        .105 

58.  Case  IV.   Given  three  sides 105 

59.  The  aid  of  logarithms  in  the  solution  of  triangles     ....  106 

60.  The  use  of  logarithms  in  Cases  I.,  II 107 

61.  Relation  between  the  sum  and  the  difference  of  any  two  sides  of  a 

triangle.    The  law  of  tangents.    Use  of  logarithms  in  Case  III.  .  108 


Xll  CONTENTS. 

ART.  PAOB 

-62.   Trigonometric  ratios  of  the  half-angles  of  a  triangle.    Use  of  loga- 
rithms in  Case  IV.         .        . 110 

63.  Problems  in  heights  and  distances   .        .        .        .        .        .        .113 

64.  Summary 115 


CHAPTER  VIII. 

LiDE  AND  Area  op  a  Triangle.    Circles  connected  with 
A  Triangle. 

65.  Length  of  a  side  of  a  triangle  in  terms  of  the  adjacent  sides  and  the 

adjacent  angles 116 

66.  Area  of  a  triangle 117 

67.  Area  of  a  quadrilateral  in  terms  of  its  diagonals  and  their  angle  of 

intersection 118 

68.  The  circumscribing  circle  of  a  triangle 119 

69.  The  inscribed  circle  of  a  triangle 119 

70.  The  escribed  circles  of  a  triangle 120 


CHAPTER  IX. 
Radian  Measure. 

71.  The  radian  defined 121 

72.  The  value  of  a  radian 122 

73.  The  radian  measure  of  an  angle.    Measure  of  a  circular  arc   .        .  123 

CHAPTER  X. 
Angles  and  Trigonometric  Functions. 

75.  Function.    Trigonometric  functions 128 

76.  Algebraical  note 129 

77.  Changes  in  the  trigonometric  functions  as  the  angle  increases  from 

0°  to  360° 131 

78.  Periodicity  of  the  trigonometric  functions 134 

79.  The  old  or  line  definitions  of  the  trigonometric  functions         .         .  135 

80.  Geometrical  representation  of  the  trigonometric  functions       .         .  137 

81.  Graphical  representation  of  functions 138 

82.  Graphs  of  the  trigonometric  functions 139 

83.  Relations  between  the  radian  measure,  the  sine,  and  the  tangent  of 

an  angle 143 


CONTENTS.  Xlll 


CHAPTER  XI. 
General  Values.    Inverse  Trigonometric  Functions. 

ART.  PAGE 

84.  General  values 146 

85.  General  expression  for  all  angles  which  have  the  same  sine     .        .  146 

86.  General  expression  for  all  angles  which  have  the  same  cosine         .  148 

87.  General  expression  for  all  angles  which  have  the  same  tangent        .  149 

88.  Inverse  trigonometric  functions ,;  .  151 

89.  Sum  and  difference  of  two  anti-tangents.    Exercises  on  inverse 

functions 152 

90.  Trigonometric  equations 164 


CHAPTER  XII. 
Miscellaneous  Theorems  and  Exercises. 

92.  Functions  of  twice  an  angle.    Functions  of  half  an  angle        .        .156 

93.  Functions  of  three  times  an  angle.    Functions  of  an  angle  in  terms 

of  functions  of  one-third  the  angle 157- 

94.  Functions  of  the  sum  of  three  angles 158 

95.  Identities .159 

q2  i03 

96.  For  an  acute  angle  of  6  radians,  cos  ^  >  1 ,  sin  ^  >  ^ ,         .    160 

4  4 

97.  One  method  of  computing  trigonometric  functions  ....     161 

98.  Trigonometry  defined.    Branches  of  trigonometry  ....    162 


APPENDIX. 

Note  A.   Historical  sketch .  165 

Note  B.  Projection  definitions  of  trigonometric  ratios  ,        .        .  .  169 

Note  C.   On  the  ratio  of  the  length  of  a  circle  to  its  diameter       .  .171 

Note  D.  On  analytical  trigonometry  and  De  Moivre's  Theorem    .  .  176 

Questions  and  Exercises  for  Practice  and  Review       .        .  .  181 

Answers  to  the  Examples 203 


PLANE    TRIGONOMETET. 


CHAPTER  I. 

LOGARITHMS:    REVIEW  OF  TREATMENT  IN  ARITH- 
METIC  AND   ALGEBRA. 

1.  There  is  a  large  amount  of  computation  necessary  in  the 
solution  of  some  of  the  practical  problems  in  trigonometry.  The 
labour  of  making  extensive  and  complicated  calculations  can  be 
greatly  lessened  by  the  employment  of  a  table  of  logarithms, 
an  instrument  which  was  invented  for  this  very  purpose  by 
John  Napier  (1550-1617),  Baron  of  Merchiston  in  Scotland,  and 
described  by  him  in  1614.  From  Henry  Briggs  (1556-1631),  who 
was  professor  at  Gresham  College,  London,  and  later  at  Oxford, 
this  invention  received  modifications  which  made  it  more  con- 
venient for  ordinary  practical  purposes."* 

Every  good  treatise  on  algebra  contains  a  chapter  on  logarithms. 
This  brief  introductory  review  is  given  merely  for  the  purpose 
of  bringing  to  mind  the  special  properties  of  logarithms  which 
make  them  readily  adaptable  to  the  saving  of  arithmetical  work. 
A  little  preliminary  practice  in  the  use  of  logarithms  will  be  of 
advantage  to  any  one  who  intends  to  study  trigonometry.  A 
review  of  logarithms  as  treated  in  some  standard  algebra  is 
strongly  recommended. 

*  The  logarithms  in  general  use  are  known  as  Common  logarithms  or  as 
Briggs' s  logarithms,  in  order  to  distinguish  them  from  another  system,  which 
is  also  a  modified  form  of  Napier's  system.  The  logarithms  of  this  other 
modified  system  are  frequently  employed  in  mathematics,  and  are  known  as 
Natural  logarithms,  Hyperbolic  logarithms,  and  also,  but  erroneously,  as 
Napierian  logarithms.  See  historical  sketch  in  article  Logarithms  (Ency. 
Brit.  9th  edition),  by  J.  W.  L.  Glaisher. 

1 


2  PLANE  'CBIGONOMETBY,  [Ch.  I 

2.  Definition  of  a  logarithm. 

If  a^  =  7V;  (1 

then  a;  is  the  index  of  the  power  to  which  a  must  he  raised  in  order 
to  equal  N. 

For  some  purposes,  this  idea  is  presented  in  these  words :  If 
a^*  =  Nj  then  x  is  the  logarithm  of  N  to  the  base  a. 

The  latter  statement  is  taken  as  the  definition  of  a  loganthmf 
and  is  expressed  by  mathematical  symbols  in  this  manner,  viz. : 

a;  =  log^^JV.  (2) 

Equations  (1),  (2),  are  equivalent ;  they  are  merely  two  different 
ways  of  stating  a  certain  connection  between  the  three  quantities 
a,  X,  N.    For  example,  the  relations 

23  =  8,    5^  =  625,    10-3  =  3-JL^  =  . 001, 

may  also  be  expressed  by  the  equivalent  logarithmic  equations, 

log2  8  =  3,    logs  625  =  4,    logio  .001  =  -  3. 

EXAMPLES. 

1.  Express  the  following  equations  in  a  logarithmic  form  : 

33  =  27,  44  =  256,  112  ^  121,  93  =  72a  73  =  343,  m^  =p, 

2.  Express  the  following  equations  in  the  exJ)onential  form  : 

log2  8  =  3,  logs  625  =  4,  logio  1000  =  3,  loga  64  =  8,  log„  P=a. 

3.  When  the  base  is  2,  what  are  the  logarithms  of  1,  2,  4,  8,  16,  32,  64, 
128,  256  ? 

4.  When  the  base  is  5,  what  are  the  logarithms  of  1,  5,  25, 125,  625,  3125  ? 
-f      5.   When  the  base  is  10,  what  are  the  logarithms  of  1, 10, 100, 1000, 10,000, 

100,000,  1,000,000,  .1,  .01,  .001,  .0001,  .00001,  .000001  ? 

6.  When  the  base  is  4,  and  the  logarithms  are  0,  1,  2,  3,  4,  5,  what  are 
the  numbers  ? 

7.  When  the  base  is  10,  between  what  whole  numbers  do  the  logarithms 
of  the  following  numbers  lie  :  8,  72,  235,  1140,  3470,  .7,  .04,  .0035  ? 

3.  Properties  of  logarithms.  Since  a  logarithm  is  the  index  of 
a  power,  it  follows  that  the  properties  of  logarithms  must  be 
derivable  from  the  properties  of  indices ;  that  is,  from  the  laws 


[. 


2.]  PROPERTIES   OF  LOGABITHMS.  3 

of  indices.     The  laws  of  indices  are  as  follows  (a,  m,  n,  being  any 
finite  quantities) : 

(1)  a^  X  a"  =^or:+V 

(2)  —  =  a-»  .  r^  =  a-«  =  a« ;  also,  ^  =  1.     .-.  a"  =  ll 

1       - 

(3)  («'")'»  =  a'"^  (4)    ■^^={a^y  =  a\ 

Let  M=a"',  whence,  log^M^m-,  (1) 

and  let  JV=  a%  whence,  log«  N=n.'  (2) 

It  follows  that 

MN  =  »*"+" ;  whence,  loga  ilOT  =  m  +  n  =  loga  ilf  +  loga  N,  (3) 

[If  P=aP,  then  log«P  =  ^,  iJf^P  =  a'"+"+^ ;    . 
whence,    log^  MNP  =  m -\- n -\- p —  \og^  M  +  log«  iV  +  log^  P.  ] 

Also,  ^  =  ^  =  a-«; 

whence,  loga  ^  =  m  -  n  =  loga  M  -  loga  N»  (4) 

Also,  3P  =  (a"*)*"  =  a'"'' ;  whence,  log<j  M^  =  rm  =  r  log^  ilf .    (5) 

1  Wl 

Also,  ■\/JW^=  (a"*)''  =  a^ ; 

^ -     1  1 

whence,  loga  V -M"  =  -  .  m  =  -  loga  M.  (6) 

The  results  (3)-(6)  state  the  properties,  or  are  the  laws  of  loga- 
rithms.    They  may  be  expressed  in  words  as  follows : 

(1)  The  logarithm  of  the  product  of  any  number  of  factoids  is  equal 
to  the  sum  of  the  logarithms  of  the  factors. 

(2)  TJie  logarithm  of  the  quotient  of  two  numbers  is  equal  to  the 
logarithm  of  the  numerator  diminished  by  the  logarithm  of  the 
denominator. 

(3)  The  logarithm  of  the  Hh  power  of  a  number  is  equal  to  r  times 
the  logarithm  of  the  number. 


4  PLATTE   TIUGONOMETBY,  [C 

(4)  The  logarithm  of  the  rth  root  of  a  number  is  equal  to  -th  of 
the  logarithm  of  the  number. 

Hence,  if  the  logarithms  {i.e.  the  exponents  of  powers)  of  num- 
bers be  used  instead  of  the  numbers  themselves,  then  the  opera- 
tions of  multiplication  and  division  are  replaced  by  those  of  addition 
and  subtraction,  and  the  operations  of  raising  to  powers  and  extract- 
ing roots,  by  those  of  multiplication  and  division. 

4.  Common  system  of  logarithms.  Any  positive  number  except 
1  may  be  chosen  as  the  base ;  and  to  the  base  chosen  there  cor- 
responds a  set  or  system  of  logarithms.  In  the  common  or  deci- 
mal system  the  base  is  10,  and,  as  will  presently  appear,  this 
system  is  a  very  convenient  one  for  ordinary  numerical  calcula- 
tions.* In  what  follows,  the  base  10  is  not  expressed,  but  it  is 
always  understood  that  10  is  the  base.  The  logarithm  of  a  num- 
ber in  the  common  system  is  the  answer  to  the  question:  "Wliat 
power  of  10  is  the  number  f  " 

Since 

l^W,     10  =  10S      100=102,      1000= W,      10000=10*,.-, 
it  follows  that 

logl=  0,loglO=  l,loglOO=  2,logl000=  3,logl0000=  4,  .... 
This  also  shows  that  the  logarithms  of  numbers 

between      1  and      10  lie  between  0  and  1,  ] 

between    10  and    100  lie  between  1  and  2,  I  (1) 

between  100  and  1000  lie  between  2  and  3,  and  so  on.  J 

For  example, 

9  =  10-^''^\         247  =  102-39270^  1453  =  W-^^"" ; 

or      log  9  =  .95424,  log  247=    2.39270,  log  1453  =   3.16227. 

Most  logarithms  are  incommensurable  numbers.  (See  Art.  9.) 
The  decimal  part  of  the  logarithm  is  called  the  mantissa,  the 

*  The  base  of  the  natural  system  of  logarithms  is  an  incommensurable 
number,  which  is  always  denoted  by  the  letter  e  and  is  approximately  equal 
to  2. 7182818284. 


4,  5.]  COMMON  LOGARITHMS.  5 

integral   part  of  the  logarithm   is  called  the   index  or  charac- 
teristic. 

The  two  great  advantages  of  the  common  system,  as  will  now 
be  shown,  are : 

(1)  The  characteristic  of  a  logarithm  can  he  written  on  mere 
inspection  ; 

(2)  The  position  of  the  decimal  point  in  a  number  affects  the  char- 
acteristic alone,  the  mantissa  being  always  the  same  for  the  same 
sequence  of  figures. 

Since         .1=    yV   =  lO'^,        .01=   ^   =10-^, 

•001  =  TT^W  =  10-^    .0001  =  ^-.^  =  10-^  ..., ' 
it  follows  that 
log  .1  =  - 1,  log  .01  =  -  2,  log  .001  =  -  3,  log  .0001  =  -  4,  etc.  (2) 

From  (1)  and  (2)  comes  the  following  rule  for  finding  the  char- 
acteristic : 

-  Wlien  the  number  is  greater  than  1,  the  characteristic  is  positive 
and  is  one  less  than  the  number  of  digits  to  the  left  of  the  decimal 
point;  when  the  number  is  less  than  1,  the  characteristic  is  negative, 
and  is  one  more  than  the  number  of  zeros  between  the  decimal  point 
and  the  first  significant  figure. 

When  a  change  is  made  in  the  position  of  the  decimal  point  in  a 
number,  the 'value  of  the  number  is  changed  by  some  integral 
power  of  10.  Its  logarithm  is  then  changed  by  a  whole  number 
only,  and,  consequently,  its  mantissa  is  not  affected.     For  example, 

25.38  =  2538  x  10-^,         2538000  =  2538  x  IQ^ ; 

and  hence,  log  25.38  =  log  2538  -  2,  log  2538000  =  log  2538  +  3. 

Accordingly,  it  is  necessary  to  put  only  the  mantissas  of  se- 
quences of  integers  in  the  tables. 

5.  Negative  characteristics.  In  common  logarithms  the  mantissa 
is  ahvays  kept  positive.  Thus,  for  example,  log  25380  =  4.40449 ; 
log  .002538  =  log  ToWAo  =  log  2538  -  log  1000000  =  3.40449  -  6 
=  -  3  +  .40449.     (Never  put  -  2.59551.) 


6  PLANE   TRIGONOMETRY.  [Ch.  1. 

This  logarithm  is  usually  written  3.40449,  in  order  to  show  that 
the  minws  sign  affects  the  characteristic  alone.  In  order  to  avoid 
the  use  of  negative  characteristics,  10  is  often  added  to  the  loga- 
rithm and  —  10  placed  after  it. 

Thus  3.40449  is  written  7.40449  - 10. 

The  second  form  is  more  convenient  for  purposes  of  calculation. 

Special  care  is  necessary  in  dealing  with  logarithms  because  of 
the  fact  that  the  mantissa  is  always  positive,  while  the  character- 
istic may  be  either  positive  or  negative.  Some  typical  examples 
involving  negative  characteristics  are  given  below. 

Addition  Subtraction  Multiplication 

/  1.  3.27412  2.  3.27412        i.e.  7.27412  -  10  3.  9.83471  -  10 

4.51459  4.51459  4.51459  2 

1.78871  2.75953  - 10  19.66942  -  20  (1) 

i.e.  9.66942  -  10  (2) 

A  result  like  (1)  is  always  put  in  the  form  (2),  in  which  the 
number  placeJ  after  the  logarithm  is  —  10. 
Ex.  3  may  also  be  worked  thus : 

(-  1  +  .83471)  X  2  =  -  2  -f  1.66942  =  1.66942. 

4.  Division.     3.27412  --  4  =  (37.27412  _  40)  ^  4  =  9.31853.-  10. 

As  in  Ex.  3  care  is  taken  that,  finally,  the  number  after  the  log- 
arithm be  — 10. 

5.  2.34175  -  5  =  (48.34175  -  50)  --  5  =  9.66835  -  10. 

6.  4.74752  x  2  =  I:i9504  ^  23.49504-30  ^  ^  gg^gg  _  ^^ 

3  3  3 

The  method  of  finding  the  logarithms  in  the  tables  when  the 
numbers  are  given,  and  the  way  to  find  the  numbers  when  the 
logarithms  are  given,  are  usually  explained  in  connection  with 
the  tables  of  logarithms. 

6.  Exercises  in  logarithmic  computation.  On  looking  at  the 
laws  of  logarithms,  (3)-(6),  Art.  3,  it  is  apparent  that  logarithms 
cannot  assist  in  the  operations  of  addition  and  subtraction.  Log- 
arithms are  of  no  service  in  computing  expressions  of  the  forms 


e.]  EXEBCtSES.  7 

M-\-  N,  31—  N.    An  expression  is  said  to  be  adapted  to  logarithmic 
computation  when  it  is  expressed  by  means  of  factors  only.     Thus, 


1 

a"  .  If 

is  adapted  to  logarithmic  ( 

computation, 

^^^0  +  6-20^+19 
7a-5b 

is  not 

;. 

EXAMPLES. 

1. 

--ii- 

Let  B  = 

6837 
4341 

Then  log  B  =  log  6837 

-  log  4341. 

log  6837 

=  3.83487 

log  4341 

=  3.63759 

.-.  logii? 

=  0.19728 

.-.  i?  =  1.575. 

2. 

Find  V.005. 

- 

] 

Let  B  =  V.005. 

Then  log  B  = 

I6g  (.005)*  -  i  log  .005  -  3- 6^897 

= 

17.69807  -^ 
.07071> 

=  8.84948  -  10  =  2.84948» 

.-.  B  = 

3.    Find  V742  x  .0769. 
Let  B  be  the  value.    Then  log  B  =  log  v'742  x  .0769  =  log  (742  x  .0769)^ 
=  ^  log  (742  X  .0769)  =  |  [log  742  +  log  .0769]. 

log  742  =  2.87040 

log  .0769  =  2.88593 

Dividing  by  5,  5 1   1.75633 

.-.  \ogB=    .35126  ,-.  i2  =  2.245. 


>'350  X  249  V350  x  249 ^ 


Let  B  be  the  value.    Then  log  i?  =  ^  (log  456  +  log  372  -  log  350  -  log  249). 

log  456  =  2.65896,  log  350  =  2.54407 

log  372  =  2.57054,  log  249  =  2.39620 

5.22950  4.94027 

4.94027 


Dividing  by  2,  2 1  .28923 

.-.  \ogB=    .14402     (See  Art.  9,  Note  1.) 

.-.  i2=  1.395. 


8  PLANE  TRIGONOMETRY,  [Ch.  I. 

5.   Find  the  value  of  x  in  34*  =  19. 

Since  .  34^  =  19, 

log  34*  =  log  19, 

X  log  34  =  log  19, 

^     log  19     1.27875       QQ.oQ  1 

^=i^  =  i:53148=-''''''"^^^^y- 

_    6.  Find  the  value  of  (a)  ^^,    (6)  ?^^,     (c)  ^^^,    {d) 


267      ^  "  315.2      ^  '  69.83      ^  '  .4231 

7.  Find  the  value  of  (a)  ^^'^  ><  ^•^'^\  (&)  8-97  x  6.36  95.83  x  76.49 

^  ^        674.2       '  ^  ^         7.84       '  ^  ^         82.97 

8.  Find  the  value  of  (a)  \/63,        (6)  V630,        (c)  VoF,        (d)  v/Tes, 


(e)  V.063,         (/)  •V/.0063. 
9.   Find  the  value  of  V63.42  x  74.95,    V6.35  x  10.87,    \/l4.21  x  17.29. 


10.   Find  the  value  of -J^^-Qx  72- 1\   J31-21  x  41.7         /4L 
>'7.81  X    6.95     \ll.39x  15.71     >'73 


7  X  85.6 


73.4x97.8 

3 

11.   Find  the  value  of  2.5637TT  12.    [—rY     ,  b  W  V  X 

13.   Find  x  from  the  equations  : 

(a)     3*  =  35,  (6)     5*  =  70,  (c)  10*  =  36, 

((^)  10*  =  127,  (c)  10*  =  765,  (/)  10*  =  1364. 


CHAPTER  11. 

TRIGONOMETRIC   RATIOS  OF  ACUTE  ANGLES. 

7.  The  name  Trigonometry  is  derived  from  two  Greek  words 
wMch  taken  together  mean  ^  I  measure  a  triangle/  *  At  the  pres- 
ent time  the  measurement  of  triangles  is  merely  one  of  several 
branches  included  in  the  subject  of  trigonometry.  The  more  ele- 
mentary part  of  trigonometry  is  concerned  with  the  calculation  of 
straight  and  circular  lines,  angles,  and  areas  belonging  to  figures 
on  planes  and  spheres.  It  consists  of  tjvo  sections,  viz.  Plane 
Trigonometry  and  Spherical  Trigonometry.  Elementary  trigonom- 
etry has  many  useful  applications,  for  instance,  in  the  measure- 
ment of  areas,  heights,  and  distances.  An  acquaintance  with  its 
simpler  results  is  very  helpful,  and  sometimes  indispensable,  in 
even  a  brief  study  of  such  sciences  as  astronomy,  physics,  and 
the  various  branches  of  engineering.  Some  modern  branches  of 
trigonometry  require  a  knowledge  of  advanced  algebra.  Their 
results  are  used  in  the  more  advanced  departments  of  mathe- 
matics and  in  other  sciences.  This  work  considers  only  the  sim- 
pler portions  of  trigonometry,  and  shows  some  of  its  applications. 

The  truths  of  elementary  trigonometry  are  founded  upon  geom- 
etry, and  are  obtained  and  extended  by  the  help  of  arithmetic 
and  algebra.  A  knowledge  of  the  principal  facts  of  plane  geome- 
try, and  the  ability  to  perform  the  simpler  processes  of  algebra, 
are  necessary  on  beginning  the  study  of  plane  trigonometry. 
Instruments  for  measuring  lines  and  angles,  and  accuracy  in 
computation  are  required  in  making  its  practical  applications. 

8.  Ratio.  Measure.  On  entering  upon  the  study  of  trigo- 
nometry it  is  very  necessary  to  have  clear  ideas  concerning  the 
terms  ratio  and  incommensurable  numbers  as  explained  in  arith- 
metic and  algebra,  for  these  terms  play  a  highly  important  part 

*  See  historical  sketch,  p.  165. 
9 


10  PLANE  TRIGONOMETRY, 


ill  the  subject.  The  study  begins  with  an  explanation  of  certain 
ratios  which  are  used  in  it  continually,  and  most  of  the  numbers 
that  appear  in  the  solution  of  its  problems  are  incommensurable. 

If  one  quantity  is  half  as  great  as  another  quantity  in  magni- 
tude, it  is  said  that  the  ratio  of  the  first  quantity  to  the  second  is 
as  one  to  two,  or  one-half.  This  ratio  is  sometimes  indicated 
thus,  1:2;  but  more  usually  it  is  written  in  the  fractional  form, 
|.  In  this  example  the  magnitude  of  the  second  quantity  is  twice 
that  of  the  first,  and  the  ratio  of  the  second  quantity  to  the  first 
is  2  : 1,  or,  adopting  the  more  usual  style,  \,  i.e.  2.  The  ratio  of 
two  quantities  is  simply  the  number  which  expresses  the  magni- 
tude of  the  one  when  compared  with  the  magnitude  of  the  other. 
This  ratio  is  obtained  by  finding  liow  many  times  the  one  quantity 
contains  the  other,  or  by  finding  what  fraction  the  one  is  of  the 
other.  It  follows  that  a  ratio  is  merely  a  pure  number,  and  that 
it  can  be  obtained  only  by  comparing  quantities  of  the  same  kind. 
Thus  the  ratio  of  the  length  3  feet  to  the  length  2  inches  is  -^, 
i.e.  18 ;  the  ratio  of  the  weight  2  pounds  to  the  weight  3  pounds 
is  f .  But  one  cannot  speak  of  the  ratio  of  3  weeks  to  10  yards, 
for  there  is  no  sense  in  the  questions :  How  many  times  does  3 
weeks  contain  10  yards  ?     What  fraction  of  10  yards  is  3  weeks  ? 

When  it  is  said  that  a  line  is  ten  inches  long,  this  statement 
means  that  a  line  one  inch  long  has  been  chosen  for  the  unit  of 
length,  and  that  the  first  line  contains  ten  of  these  units.  Thus 
the  7iumher  used  in  telling  the  length  of  a  line  is  the  ratio  of  the 
length  of  this  line  to  the  length  of  another  line  which  has  been 
chosen  for  the  unit  of  length.  The  measure  of  any  quantity,  such 
as  a  length,  a  weight,  a  time,  an  angle,  etc.,  is 

/  the  number  of  times  the  quantity  contains  \ 
I  or,  the  fraction  that  the  quantity  is  of         / 

a  certain  quantity  of  the  same  kind  which  has  been  adopted  as  the 
unit  of  measurement.  In  other  words,  the  measure  of  a  quantity 
is  the  ratio  of  the  quantity  to  the  unit  of  measurement.  For 
example,  if  half  an  inch  is  the  unit  of  length,  then  the  measure  of 
a  line  8  inches  long  is  16 ;  if  a  foot  is  the  unit  of  length,  then  the 
measure  of  the  same  line  is  f ;  if  a  second  is  the  unit  of  time,  then 
the  measure  of  an  hour  is  3600 ;  if  an  hour  is  the  unit  of  time, 
then  the  measure  of  a  second  is  3  ^nriT' 


8.]  RATIO.      MEASURE.  11 

If  two  quantities  have  a  common  unit  of  measurement,  then 
their  ratio  is  the  ratio  of  their  measures.  For  example,  1 
pound  being  taken  as  the  unit  of  weight,  the  ratio  of  a  weight  3 
pounds  to  a  weight  7  pounds  is  -f-,  which  is  also  the  ratio  of  the 
measures  3  and  7.  In  general,  if  a  quantity  P  contains  m  units, 
and  a  quantity  Q  contains  n  units  of  the  same  kind  as  is  used  in 
the  case  of  P,  then  the  ratio 

quantity  P  _m  units  _  m^ 
quantity  Q      n  units       n 

The  last  fraction  —  is  the  ratio  of  the  numbers  m  and  n,  which 
n 

are  the  measures  of  the  quantities  P  and  Q  respectively. 

EXAMPLES. 

1.  What  is  the  ratio  of  each  of  the  following  lengths  to  an  inch,  viz., 

8  in.,  2  ft.,  3  ft.  6  in.,  1.5  yd.,  20  yd.,  a  yd.,  h  ft.,  c  in.? 

2.  What  is  the  ratio  of  eacli  of  the  following  lengths  to  a  yard,  viz., 
6  yd.,  3.75  yd.,  8  ft.,  2  ft.  6  in.,  10  in.,  5  in.,  a  yd.,  &  ft.,  c  in.  ? 

3.  What  is  the  measure  of  each  of  the  following  lengths,  when  a  foot  is 
the  unit  of  length,  viz.,  1.5  mi.,  17  yd.,  3  yd.  2  ft.,  8.5  ft.,  2  ft.  6  in., 

9  in.,  2  in.,  a  yd.,  h  ft.,  c  in.  ? 

4.  What  is  the  measure  of  each  of  the  following  lengths,  when  3  in. 
is  the  unit  of  length,  viz.,  2.5  yd.,  1.5  ft.,  8  in.,  a  yd.,  h  ft.,  c  in.  ? 

5.  Express  the  ratio  of  2.5  mi.  to  10  yd. ;  and  the  ratio  of  2\  in.  to 
3iyd. 

6.  Compare  the  ratio  of  a  foot  to  a  yard  with  the  ratio  of  a  square  foot 
to  a  square  yard. 

7.  What  is  the  unit  of  measurement  in  each  of  the  following  cases  :  when 
the  measure  of  2  ft.  is  4,  of  1  yd.  is  72,  of  .5  in.  is  4,  of  2.5  ft.  is  .25  ? 

N.B,  The  following  examples  will  he  used  again  for  purposes  of  illustra- 
tion. The  student  is  advised  to  draw  figures  neatly  and  accurately  and  to 
preserve  the  results  carefully. 

8.  In  a  right-angled  triangle  the  base  is  6  ft.  and  the  hypotenuse  10  ft. 
What  is  the  perpendicular  ?     Calculate  the  following  ratios,  viz. : 


perpendicular 

base 

perpendicular 

hypotenuse 
base 

hypotenuse 

hypotenuse 
base 

base 
hypotenuse 

perpendicular 

perpendicular 

12  PLANE  TRIGONOMETRY,  [C 

What  are  these  ratios  in  a  triangle  whose  base  is  6  in.,  and  hypotenuse 
10  in.?  What  are  they  when  the  base  is  6  yd.,  and  the  hypotenuse 
10  yd.  ?  When  the  base  is  6  mi.,  and  the  hypotenuse  10  mi.  ?  When 
the  base  is  12  ft.,  and  the  hypotenuse  20  ft.  ?  When  the  base  is  3  in., 
and  the  hypotenuse  5  in.  ?  Compare,  if  possible,  the  angles  in  these 
triangles. 

9.  In  a  right-angled  triangle  whose  base  is  35  ft.  and  perpendicular 
12  ft.,  what  is  the  hypotenuse  ?  For  this  triangle  calculate  the  ratios 
specified  in  Ex.  8.  Calculate  these  ratios  for  a  triangle  whose  base  is 
70  yd.,  and  perpendicular  24  yd.  Compare,  if  possible,  the  angles  in  these 
triangles. 

10.  Calculate  these  ratios  for  the  triangle  whose  hypotenuse  is  29  ft. ,  and 
perpendicular  21  ft.  ;  for  the  triangle  whose  hypotenuse  is  2.9  in.,  and 
perpendicular  2.1  in.     Compare,  if  possible,  the  angles  in  these  triangles. 

9.  Incommensurable  quantities.  Approximations.  If  the  side  of 
a  square  is  one  foot  in  length,  then  the  length  of  a  diagonal  of  the 
square  is  V2  feet.  Thus  the  ratio  of  the  diagonal  to  the  side  is 
V2,  a  number  which  cannot  be  expressed  as  the  ratio  of  two 
whole  numbers.  Two  quantities  whose  ratio  can  be  expressed  by 
means  of  two  integers  are  said  to  be  commensurable  the  one  with 
the  other;  when  their  ratio  cannot  be  so  expressed,  the  one  quan- 
tity is  said  to  be  incommensurable  with  the  other.  For  example, 
the  diagonal  of  a  square  is  incommensurable  with  the  side,  and 
the  length  of  a  circle  with  its  diameter.^*  The  quantities  in  the 
examples,  Art.  8,  are  commensurable.  Numbers  such  as  V2,  V4, 
VlO  are  incommensurable  with  unity,  and  their  values  cannot  be 
found  exactly.  Their  values,  however,  can  be  found  to  two,  to 
three,  to  four,  in  fact,  to  as  many  places  of  decimals  as  one 
please.  The  greater  the  number  of  places  of  decimals,  the  more 
nearly  will  the  calculated  values  represent  the  true  values  of  the 
numbers.  In  other  words,  the  values  of  incommensurable  num- 
bers can  be  found  approxiinately  ;  and  the  degree  of  approximation 
(that  is,  the  nearness  to  the  exact  values)  will  depend  only  on  the 
carefulness  and  patience  of  the  calculator.  In  practical  problems 
there  frequently  is  occasion  for  the  exercise  of  judgment  as  to  the 
degree  of  approximation  that  is  necessary  and  sufficient.  For 
example,  in  calculating  a  length  in  inches  in  ordinary  engineer- 

*  See  Appendix,  Note  C. 


9.]  APPBOXIMATIONS.  13 

ing  work  there  is  no  need  to  go  beyond  the  third  place  of  deci- 
mals, for  engineers  are  satisfied  when  a  measurement  is  correct 
to  within  -^^  of  an  inch.  As  a  rule  the  results  obtained  in  practical 
problems  in  mathematics  are  only  approximate  and  not  exact. 
There  are  two  reasons  for  this :  first,  the  data  obtained  by  actual 
measurement  can  only  be  approximate,  however  excellent  the 
instruments  used  in  measuring  may  be,  and  however  skilled  and 
careful  is  the  person  who  does  the  measuring ;  second,  most  of 
the  numbers  used  in  the  subsequent  computations  are  incommen- 
surable. 

The  examples  at  the  end  of  this  article  are  intended  to  bring  out  more 
clearly  the  idea  of  an  approximate  result.  The  answers  are  to  be  calculated 
to  three  places  of  decimals.  It  is  advisable  to  compare  the  values  calculated 
to  three  places  of  decimals  with  the  values  calculated  to  two  places  of  deci- 
mals, and  to  note  the  difference  between  them.  The  following  facts  are  sup- 
posed to  be  known  and  will  be  taken  for  granted. 

(a)  In  a  right-angled  triangle  the  square  of  the  measure  of  the  hypotenuse 
is  equal  to  the  sum  of  the  squares  of  the  measures  of  the  other  two  sides. 

(&)  The  ratio  of  the  length  of  any  circle  to  its  diameter  is  a  number  which 
is  the  same  for  all  circles.*  The  exact  value  of  this  ratio  is  incommensurable 
and  is  always  denoted  by  the  symbol  tt  (read;9i).t  The  approximate  values 
commonly  used  for  tt  are  3.1416,  3.14159,  fff  (i.e.  3.1415929  -.•),  ¥  (i.e. 
3.142857)  ;  of  these  values  the  last  is  the  least  accurate,  but  it  is  accurate 
enough  for  many  practical  purposes. 

(c)  The  length  of  a  circle  of  radius  r  is  2  irr  [by  (6)]  ;  and  the  enclosed 
area  is  irr^. 

Note  1.  If  a  number  be  calculated  to  three  or  more  places  of  decimals, 
then  the  closest  approximation  to,  say,  two  places  of  decimals  is  obtained  by 
leaving  the  number  in  the  second  place  of  decimals  unchanged  when  the 
number  in  the  third  place  is  less  than  5,  and  by  increasing  the  number  in  the 
second  place  by  unity  when  the  number  in  the  third  place  is  greater  than  5 
or  5  followed  by  numbers;  thus,  e.g.,  3.72  for  3.724,  3.73  for  3.7261  and 

*  This  ratio  and  facts  (c)  are  considered  in  Note  C,  Appendix.  The  read- 
ing only  requires  a  knowledge  of  elementary  geometry. 

t  This  symbol  is  the  initial  letter  of  periphereia,  the  Greek  word  for  cir- 
cumference. Its  earliest  appearances  to  denote  this  ratio  are  in  Jones's 
Synopsis  Palmarioriim  Mathesos,  London,  1706,  and  in  the  Introductio  in 
analysin  infinitorum.,  published  in  1748  by  Leonhard  Euler  (1707-1783),  a 
native  of  Switzerland,  who  was  one  of  the  greatest  mathematicians  of  his 
time. 


14  PLANi:  fRIGONOMETRY.  [Ch.  I 


3.7257.  When  the  number  in  the  third  place  is  5  and  this  is  followed  by 
zeros  only,  the  number  in  the  second  place  is  unchanged  if  it  is  even,  and  is 
increased  by  unity  if  it  is  odd  ;  thus,  e.g.^  3.78  for  3.775,  3.78  for  3.785.  In 
a  series  of  calculations  the  errors  made  by  following  this  rule  tend  to  balance 
one  another. 

Note  2.  A  quantity  measured  to  two  places  of  decimals  is  correct  to  the 
hundredth  part  of  the  unit  employed,  and  a  quantity  measured  to  three  places 
is  correct  to  the  thousandth  part  of  the  unit.  For  example,  the  length  of  a 
circle  of  10  feet  diameter  is  31.4159  .  .  .  feet.  For  this  length  31.416  or 
31.42  maybe  taken;  the  former  result  differs  from  the  true  result  by  less 
than  one-thousandth  of  a  foot,  the  latter  by  less  than  one-hundredth. 

EXAMPLES. 

1.  A  finds  the  square  root  of  3  correctly  to  two  places'of  decimals,  and  B 
to  three.  How  much  closer  than  A  does  B  come  to  the  exact  value  of  the 
square  root  of  3  ? 

2.  A  circle  is  50  ft.  in  diameter.  In  calculating  its  length  A  takes  3.1416 
as  the  ratio  of  the  length  of  a  circle  to  the  diameter,  B  takes  3.14159,  and  C 
takes  22  :  7.     What  are  the  differences  (in  inches)  between  their  results  ? 

3.  The  radius  of  a  circle  is  49.95  ft.  How  nearly  will  a  person  come  to 
the  length  of  the  circle  if  he  assumes  the  radius  to  be  50  ft.  ?  [In  this  and 
the  following  example  take  tt  =  22  :  7.] 

4.  It  is  known  that  the  diameter  of  a  certain  circle  does  not  differ  from 
100  ft.  by  more  than  2  in.  What  will  be  the  outside  limits  of  the  error 
made  in  calculating  the  area  when  the  diameter  is  taken  as  100  ft.? 

5.  Find  the  difference  between  the  calculations  of  the  numbers  of  revolu- 
tions per  mile  made  by  a  50-in.  bicycle,  for  tt  =  22  :  7  and  ir  =  3.1416. 

6.  A  lot  is  75  ft.  by  200  ft.  Find  the  diagonal  distance  across  the  lot 
correctly  to  within  a  tenth  of  an  inch. 

7.  Find  the  height  of  an  equilateral  triangle  whose  side  is  20  yd. 

8.  The  side  of  an  isosceles  triangle  is  40  ft.  and  the  base  is  30  ft. ;  find 
the  height. 

9.  What  is  the  length  of  the  diagonal  of  a  square  whose  side  is  20  ft.  ? 

10.  What  is  the  length  of  the  side  of  a  square  whose  diagonal  is  20  ft.  ? 

N.  B.  The  following  examples  will  be  used  again  for  purposes  ofillnstra- 
tion.  The  student  is  advised  to  draw  figures  and  to  preserve  the  results  icith 
those  of  Exs.  8,  9,  10,  Art.  8. 

11.  (a)  In  a  right-angled  triangle  the  hypotenuse  is  12  ft.  and'the  base  is 
6  ft.  ;  calculate  the  ratios  specified  in  Ex.  8,  Art.  8. 

(6)  What  are  these  ratios  when  the  lengths  in  (a)  are  taken  twice,  three 
times,  one-half  as  great  ?     Compare,  if  possible,  the  angles  in  these  triangles. 


»Tr™l 


10.]  DRAWING   TO   SCALE,  15 

12.  (a)  In  a  right-angled  triangle  the  base  is  8  in.  and  the  perpendicu- 
lar 12  in.  ;  calculate  the  ratios  specified  in  Ex.  8,  Art.  8. 

(&)  What  are  these  ratios  when  the  lengths  in  (a)  are  taken  one-third, 
and  four  times  as  great  ?    Compare,  if  possible,  the  angles  in  these  triangles. 

13.  (a)  In  a  right-angled  triangle  the  hypotenuse  is  35  yd.  and  the  per- 
pendicular is  15  yd.  ;  calculate  the  ratios  specified  in  Ex.  8,  Art.  8. 

(h)  What  are  these  ratios  when  the  lengths  in  (a)  are  taken  four  times, 
six  times,  one-fifth  as  great  ?  Compare,  if  possible,  the  angles  in  these 
triangles. 

10.  Linear  measure.  Drawing  to  scale.  Direct  measurement  by 
means  of  drawings.  Various  systems  of  linear  nieasurement  are 
described  in  arithmetic.  The  system  mostly  used  in  English- 
speaking  countries  is  that  in  which  length  is  given  in  miles, 
yards,  feet,  or  inches.  The  system  which  is  in  common  use  on 
the  continent  of  Europe,  and  which  is  mainly  employed  in 
scientific  measurements  throughout  the  world,  is  the  metric 
system.  In  this  system  lengths  are  given  in  centimetres,  metres, 
etc.,  the  centimetre  being  a  hundredth  part  of  a  metre.  A  metre 
is  equal  to  39.37  .  .  .  inches.* 

Drawing  to  scale.  It  is  often  desirable  to  have  a  drawing  on 
paper  which  shall  serve  to  give  an  accurate  idea  of  the  relations 
of  certain  lines  and  positions.  Maps  and  architects'  plans  are 
familiar  examples  of  such  drawings.  In  a  map  an  inch  may 
represent  1  mile,  10  miles,  100  miles,  500  miles,  and  so  on,  accord- 
ing to  the  scale  on  which  the  map  is  made ;  in  a  building  plan 
an  inch  may  represent  10  feet,  12  feet,  and  so  on.  The  operation 
of  drawing  on  paper  lines  that  shall  be  a  half,  a  quarter,  a  tenth, 
a  thousandth,  etc.,  part  of  the  actual  length  of  given  lines,  is 
called  drawing  to  scale.  In  many  cases  the  drawings  of  objects 
cannot  be  made  full  size ;  for  instance,  the  map  of  a  town,  the 
floor  plan  of  a  church ;  these  are  drawn  to  a  reduced  scale.  In 
other  cases  the  drawings  are  made  larger  than  the  actual  objects, 
for  instance,  the  drawings  of  the  minute  things  that  live  in  a 
drop  of  water,  the  drawings  of  the  various   parts  of   a  flower; 

*  The  metric  system  has  the  great  advantage  of  being  a  decimal  system. 
At  the  present  time  committees  of  scientific  societies  in  England  and  America 
are  working  to  have  the  common  system  replaced  by  the  metric. 


16  PLANE  TRIGONOMETRY.  [Ch.  II. 

these  are  drawn  to  an  enlarged  scale.  When  a  drawing  is  made  to 
scale,  the  scale  should  always  be  indicated  on  it.  This  may  be  done 
in  various  ways.     Thus  a  mere  statement  may  be  made,  e.g., 

1  inch  to  10  feet;  / 

or,  the  scale  may  be  indicated  by  a  fraction  which  gives  the  ratia 
of  any  line  in  the  drawing  to  the  actual  line  represented.  The 
scale  can  also  be  shown  graphically  by  means  of  a  specially  marked 
line.  Both  the  latter  methods  are  illustrated,  for  instance,  on 
the  map  of  the  Kingdom  of  Saxony  in  Tlie  Times  Atlas : 


1:870000. 

10 

English  Miles  (69.16  ^  1° ) 

Pig.  1. 

The  scale  should  be  expressed  fractionally,  that  is,  by  expresjy- 
ing  the  ratio  of  a  line  in  the  drawing  to  the  actual  line  repre- 
sented. Thus  in  the  first  example  above  the  scale  is  1 :  120 ;  in 
the  second  the  scale  is  1 :  870000. 

When  a  drawing  is  made  to  scale,  the  distance  between  tiuo  objects 
can  be  measured  directly,  by  merely  measuring  the  distance  between 
their  corresponding  points  on  the  drawing.  For  instance,  if  1  inch 
represents  120  feet,  then  2.5  inches  represents  300  feet.  Another 
example :  On  the  map  of  Saxony  referred  to  above,  the  distance 
between  Leipzig  and  Dresden  is,  approximately,  4i  inches,  and  4^ 
inches  x  870000  gives  about  62  miles  as  the  distance  in  an  air- 
line between  these  cities.  This  method  of  finding  distance  can  be 
used  in  solving  many  of  the  problems  in  trigonometry.*  To  find 
the  length  of  the  representative  line  in  the  drawing  when  the 
scale  and  the  actual  length  are  given,  is  an  exercise  in  simple 
proportion ;  so,  also,  to  find  the  actual  length  of  a  line  when  the 
scale  and  the  length  of  the  representative  line  are  known. 

*  This  is  one  of  the  methods  which  will  be  employed  in  this  book  in  prob- 
lems involving  distance.  Proficiency  in  drawing  will  be  very  helpful  to  the 
student. 


10.]  EXAMPLES.  17 

EXAMPLES. 

1.  When  an  inch  represents  10  ft.,  how  long  must  the  lines  be  that  will 
represent  3  in.,  6  in.,  1  ft.,  2  ft.,  5  ft.,  15  ft.,  7.5  ft.,  30  ft.,  40ft.,  55  ft.? 
What  is  the  scale  ? 

2.  When  an  inch  represents  5  yd. ,  how  long  must  the  lines  be  tiiat  will 
represent  2  yd.,  4  yd.,  7  yd.,  11  yd.,  3  yd.  2  ft.,  4  yd.  1  ft.  8  in.  ?  What 
is  the  scale  ? 

3.  When  an  inch  represents  150  ft.,  what  distances  are  represented  by 
I  in.,  I  in.,  i  in.,  If  in.,  2|  in.,  4.8  in.,  5.3  in.  ?     What  is  the  scale  ? 

4.  When  an  inch  represents  10  mi.,  what  distances  are  represented  by 
3  in.,  7  in.,  ^  in.,  ^  in.,  3 J  in.  ?  What  lengths  on  the  drawings  will  represent 
7  mi.,  18  mi.,  25  mi.  ?    What  is  the  scale  ? 

5.  What  are  the  scales  when  1  in.  represents  100  ft.,  ^  in.  represents  a 
mile,  I  in.  represents  20  ft.,  ^^  in.  represents  15  yd.,  1  in.  represents  1  mi., 
10  mi.,  100  mi.  ? 

6.  Draw  to  a  scale  1 :  240  (20  ft.  to  the  inch)  the  circles  in  Exs.  2,  3,  4,  5, 
Art.  9. 

7.  On  a  map  in  Baedeker's  Guide  to  Paris  the  distance  between  the 
nearest  corners  of  the  Eiffel  Tower  and  Notre  Dame  Cathedral  is  7f  in. 
What  is  the  distance  between  those  points,  the  map  being  drawn  to  a  scale 
1  :  20000  ? 

8.  Make  the  comparison  of  angles  asked  for  in  Exs.  8,  9,  10,  Art.  8  ;  Exs. 
11,  12,  13,  Art.  9. 

Suggested  Exercises.  Make  drawings  to  scale  of  the  floor  plan  of  a 
dwelling  house,  of  some  other  building,  of  some  grounds.  Find  the  distances 
between  various  points,  such  as  diagonally  opposite  corners,  by  making 
measurements  in  the  drawing  and  applying  the  scale.  Compare  the  results 
obtained  in  this  way  with  the  results  obtained  by  other  methods.  Other 
methods  that  may  be  used  are :  (1)  making  an  off-hand  estimate  of  the  dis- 
tance ;  (2)  actually  measuring  the  distance  by  "pacing  off"  or  by  using  a 
rule  or  tape  line  ;  (3)  making  a  computation.  Let  the  student,  from  his 
own  experience,  form  a  judgment  as  to  which  of  the  four  methods  referred  to 
is  the  easiest,  and  which  the  more  exact.  Find  the  air-line  distances  be- 
tween places  by  measuring  the  distances  between  them  on  maps.  Several 
maps  may  by  used  so  as  to  have  a  variety  of  scales. 

Note.  The  word  scale  also  has  another  meaning  in  drawing  and  meas- 
urement. Engineers  and  draughtsmen  use  various  kinds  of  rules  called 
scales.  The  faces  of  these  rules  contain  different  numbers  of  divisions  to  an 
inch,  one  10  divisions,  one  20,  one  30,  and  so  on  ;  and  generally,  one  inch  on 
each  face  is  subdivided  so  that  a  small  fraction  of  an  inch  may  be  set  off  or 
read.     Some  paper  scales  are  on  the  protractor  inserted  in  this  book. 


18  PLANE  TRIGONOMETRY.  [Ch.  I 


» 


11.  Degree  measure.  The  protractor.  It  has  been  seen 
geometry  that:  (1)  When  one  line  is  perpendicular  to  another 
line,  each  of  the  angles  made  at  their  intersection  is  a  right  angle ; 
(2)  All  right  angles  are  equal  to  one  another.  In  some  geomet- 
rical propositions  angles  are  compared,  and  one  angle  is  shown  to 
be  greater  or  less  than  another.  But  geometry,  with  the  excep- 
tion of  a  few  cases,  does  not  show  by  exactly  how  7nuch  the  one 
angle  is  greater  or  less  than  the  other.  In  order  to  show  this, 
measurement  is  necessary ;  and  in  order  to  measure,  a  unit  angle 
of  measurement  must  be  chosen.  The  unit  of  angular  magnitude 
which  is  generally  used  in  practical  work  is  the  angle  that  is  one- 
ninetieth  part  of  a  right  angle.  This  unit  angle  is  called  a  degree. 
All  degrees  are  equal  to  one  another,  since  all  right  angles  are 
equal  to  one  another.  Each  degree  is  subdivided  into  60  equal 
parts  called  minutes,  and  each  minute  is  subdivided  into  60  equal 
parts  called  seconds.     Hence  comes  the  following  table  of  angular 

measure : 

60  seconds  =  1  minute, 
60  minutes  =  1  degree, 
90  degrees  =  1  right  angle. 

The  magnitude  of  an  angle  containing  37  degrees  and  42  minutes  and  35 
seconds,  say,  is  written  thus :  37°  42'  35",  read  37  degrees,  42  minutes,  35 
seconds.  This  system  of  measurement  is  sometimes  called  the  rectangular 
system^  sometimes  tlie  sexagesimal  system.  In  this  chapter  only  acute  angles, 
that  is,  angles  which  contain  between  0°  and  90°,  are  considered.  Chapter 
V.  considers  angles  of  all  magnitudes. 

Note  1.  An  angle  1°  is  subtended  by  1  in.  at  a  distance  4  ft.  9.3  in.,  and 
by  1  ft.  at  a  distance  57.3  ft.  An  angle  1'  is  subtended  by  1  in.  at  a  distance 
286.5  ft.,  and  by  1  ft.  at  a  distance  3437.6  ft.,  about  two-thirds  of  a  mile. 
An  angle  1"  is  subtended  by  1  in.  at  a  distance  of  nearly  3|  mi.,  by  1  ft.  at  a 
distance  a  little  greater  than  39|  mi.,  by  a  horizontal  line  200  ft.  long  on  the 
other  side  of  the  world,  nearly  8000  mi.  away.  These  facts  can  be  verified 
later.     See  Ex.  3,  Art.  83. 

Note  2.  Another  system  of  angular  measurement  was  advocated  by 
Briggs  and  other  mathematicians  (see  Art.  1),  and  was  introduced  in  France 
at  the  time  of  the  Revolution.  In  this  system,  which  is  a  decimal  o^ie  and 
called  the  centesimal  system.,  a  right  angle  is  divided  into  100  equal  parts 
called  grades.,  each  grade  into  100  equal  parts  called  minutes,  and  each 
minute  into  100  equal  parts  called  seconds.  It  has  not  been  generally 
adopted,  on  account  of  the  immense  amount  of  labour  that  would  be  necessary 
in  order  to  change  the  mathematical  tables  computed  for  the  other  system. 


|]  THE  PROTRACTOR.  19 

Note  3.  The  sexagesimal  system  (from  sexagesimus,  sixtieth)  was  in- 
vented by  the  Babylonians,  who  constructed  their  tables  of  weights  and 
measures  on  a  scale  of  60.  Their  tables  of  time  (1  day  =  24  hr.,  1  hr.  =  60 
min.,  1  min.  =  60  sec.)  and  circular  measure  have  come  down  to  the  present 
day.  It  has  been  suggested  that  their  adoption  of  the  scale  of  60  is  due  to 
the  fact  that  they  reckoned  the  year  at  360  days.  "  This  led  to  the  division 
of  the  circumference  of  a  circle  into  360  degrees,  each  degree  representing  the 
daily  part  of  the  supposed  yearly  revolution  of  the  sun  around  the  earth 
Probably  they  knew  that  the  radius  could  be  applied  to  the  circumference  as 
a  chord  six  times,  and  that  each  arc  thus  cut  off  contained  60  degrees.  Thus 
the  division  into  60  parts  may  have  suggested  itself .  .  .  .  Babylonian  science 
has  made  its  impress  upon  modern  civilization.  Whenever  a  surveyor  copies 
the  readings  from  the  graduated  circle  on  his  theodolite,  whenever  the 
modern  man  notes  the  time  of  day,  he  is,  unconsciously  perhaps,  but  unmis- 
takably, doing  homage  to  the  ancient  astronomers  on  the  banks  of  the  Eu- 
phrates."—  Cajori,  History  of  Elementary  Mathematics^  pp.  10,  11. 

Note  4.  Another  system  of  angular  measure  is  described  in  Chapter  IX. 
See  Art.  71. 

The  protractor.  The  protractor  is  an  instrument  used  for  meas- 
uring given  angles  and  laying  off  required  angles  on  paper.  Pro- 
tractors are  of  various  kinds,  of  which  the  semicircular  and  the 
full-circled  are  the  most  common.  The  degrees  are  marked  all 
round  the  edge.  A  paper  protractor  is  inserted  in  this  book  for 
use  in  solving  problems.*  In  order  to  draw  a  line  that  shall 
make  a  given  angle  with  a  given  line  at  a  given  point,  proceed 
as  follows :  Place  the  centre  of  the  protractor  at  the  given  point 
and  bring  its  diameter  into  coincidence  with  the  given  line,  keep- 
ing the  semicircle  on  the  side  on  which  the  required  line  is  to  be 
drawn;  prick  off  the  required  number  of  degrees  with  a  sharp 
pencil  or  fine  needle.  The  line  joining  the  point  thus  fixed  and 
the  given  point,  is  the  line  required.  In  order  to  measure  a  given 
angle  with  the  protractor,  place  the  centre  at  the  vertex  of  the 
angle,  and  place  the  diameter  in  coincidence  with  one  of  the 
boundary  lines  of  the  angle;  the  number  of  degrees  in  the  arc 
intercepted  between  the  boundary  lines  of  the  angle  is  the  meas- 
ure of  the  angle. 

*  A  horn  protractor  costs  about  25  cents,  and  a  small  metal  one  about  50 
cents.     One  who  is  neat  and  handy  can  make  a  paper  protractor. 


20 


PLANE  TBIGONOMETRY. 


[Ch.  II. 


Note.  Before  proceeding  further,  the  student  should  be  able  to  draw 
with  ease  a  right-angled  triangle,  having  been  given:  (a)  The  hypotenuse 
and  a  side  ;  {h)  the  two  sides  about  the  right  angle ;  (c)  the  hypotenuse 
and  one  of  the  acute  angles  ;  (d)  one  of  the  sides  about  the  right  angle  and 
the  opposite  angle  ;  (e)  one  of  the  sides  about  the  right  angle  and  the  adja- 
cent angle.  It  is  here,  taken  for  granted  that  these  problems  have  been  con- 
sidered in  a  course  in  plane  geometry  or  in  a  course  of  geometrical  drawing. 


EXAMPLES. 

N.  B.  The  student  is  advised  to  do  Exs.  1-6  carefully,  and  to  preserve  the 
results,  for  they  will  soon  be  required  for  purposes  of  illustration. 

1.  Draw  to  scale  the  triangles  considered  in  Exs,  8,  9,  10,  Art.  8,  and  Exs. 
11,  12,  13,  Art.  9,  and  measure  the  angles. 

2.  Make  drawings,  on  two  different  scales,  of  a  right-angled  triangle 
whose  base  is  20  ft.  and  adjacent  acute  angle  is  55°.  In  each  drawing  meas- 
ure the  remaining  parts  and  thence  deduce  the  unknown  parts  of  the  original 
triangle.     In  each  drawing  calculate  the  ratios  specified  in  Ex.  8,  Art.  8. 

3.  Same  as  Ex.  2,  for  a  right-angled  triangle  whose  hypotenuse  is  30  ft. 
and  angle  at  base  is  25". 

4.  Same  as  Ex.  2,  for  a  right-angled  triangle  whose  base  and  perpendicular 
are  30  ft.  and  45  ft.  respectively. 

5.  Same  as  Ex.  2,  for  a  right-angled  triangle  whose  hypotenuse  is  60  ft. 
and  base  is  45  ft. 

6.  Same  as  Ex.  2,  for  a  right-angled  triangle  in  which  the  base  is  50  ft. 
and  the  angle  opposite  to  the  base  is  40°. 

7.  What  angles  of  a  whole  number  of  degrees  can  easily  be  constructed 
geometrically  without  the  aid  of  the  protractor  ?     Make  the  constructions. 

12.  Trigonometric  ratios  defined  for  acute  angles.  The  ratios 
referred  to  at  the  beginning  of  Art.  8  will  now  be  explained  so  far 
as  acute  angles  are  concerned.     (Before  proceeding,  the  student 


Via.  2, 


12.] 


THE  TBIGONOMETRIC  RATIOS. 


21 


should  glance  over  the  work  on  Exs.  8-10,  Art.  8 ;  Exs.  11-13,  Art. 
9 ;  Exs.  1-6,  Art.  11.)  Let  A  be  any  acute  angle.  In  either  one  of 
the  lines  containing  the  angle  take  any  point  P  and  let  fall  a 
perpendicular  FM  to  the  other  line.  The  three  lines  AP,  AM, 
MP,  can  be  taken  by  twos  in  three  different  ways,  and  hence  six 
ratios  can  be  formed  with  them,  namely : 

MP     AM     MP     AM      AP^     AP 
AP      AP'    am'     MP     AM      MP 

It  is  shown  in  Art.  13  that  each  of  these  ratios  has  the  same 
value  as  in  Fig.  2,  no  matter  where  the  point  P  is  taken  on  either 
one  of  the  lines  bounding  an  angle  which  is  equal  to  A.  For  the 
sake  of  convenience  of  reference,  each  one  of  these  six  ratios  is  given 
a  particular  name  with  respect  to  the  angle  A.     Thus ; 


MP 

— — -  is  called  the      sine      of  the  angle  A ; 
AP  ^        ' 

AM 

— -—  is  called  the     cosine    of  the  angle  A ; 
AP  ^         ' 

MP  ' 

■ is  called  the   tangent  of  the  angle  A  ; 

AM 

AM 

is  called  the  cotangent  of  the  angle  A : 

MP 

AP 

— —  is  called  the     secant    of  the  angle  A ; 

AP 

is  called  the  cosecant  of  the  angle  A. 


■  ay 


These  six  ratios  are  known  as  the  trigonometric  ration  of  the 
angle  A.  According  to  the  definition  of  a  ratio  (Art.  8)  they  are 
merely  numbers.     Eor  brevity  they  are  written  sm  A,  cos  A,  tan  A, 


*  In  Chapter  V.  the  trigonometric  ratios  are  defined  for  angles  in  general. 
The  definitions  givo^iri  this  article  will  be  found  to  follow  immediately  from 
those  given  in  Art.  40. 


22  PLANE   TRIGONOMETRY.  [Ch.  J(| 

cot  A,  sec  A,  cosec  A  (or  esc  A).*  Thus  tan  A  is  read  ^'tangent 
J.,"  and  means  "the  tangent  of  the  angle  ^."  The  giving  of 
names  in  (1)  may  be  regarded  as  defining  the  trigonometric  ratios. 
Definitions  (1)  may  be  expressed  as  follows : 


AP  '        AM  '        AM 

4^  =  cos  ^,        4^=  cot  A,  ^  =cosec  A, 

AP  '         MP  '  MP  ' 


.     .     (2) 


These  definitions  can  be  given  a  slightly  different  form  which 
is  more  general,  and,  accordingly,  more  useful  in  applications.  In 
any  right-angled  triangle  AMP  (Fig.  2),  M  being  the  right 
angle,  with  reference  to  the  angle  A  let  MP  be  denoted  as  the 
opposite  side,  and  AM  as  the  adjacent  side.  Then  these  defini- 
tions take  the  form :  — 


*  The  term  sine  first  appeared  in  the  twelfth  century  in  a  Latin  translation 
of  an  Arabian  work  on  astronomy,  and  was  first  used  in  a  published  work  by 
a  German  mathematician,  liegiomontanus  (1436-1476).  The  terms  secant 
and  tangent  were  introduced  by  a  Dane,  Thomas  FincJc  (1561-1646),  in  a 
work  published  in  1583.  The  term  cosecant  seems  to  have  been  first  used  by 
Bheticus,  a  German  mathematician  and  astronomer  (1514-1576),  in  one  of 
his  works  which  was  published  in  1596.  The  names  cosine  and  cotangent 
were  first  employed  by  Edmund  Gunter  (1581-1626),  professor  of  astronomy 
at  Gresham  College,  London,  who  made  the  first  table  of  logarithms  of  sines 
and  tangents,  published  in  1620,  and  introduced  the  Gunter's  chain  now 
used  in  land  surveying.  The  abbreviations  sin,  tan,  sec,  were  first  used 
in  1626  by  a  Flemish  mathematician,  Albert  Girard  (1590-1634),  and  those 
of  cos,  cot,  appear  to  have  been  earliest  used  by  an  Englishman,  AVilliam 
Oughtred  (1574-1660),  in  his  Trigonometry,  published  in  1657.  Tliese  con- 
tractions, however,  were  not  generally  adopted  until  after  their  reintroduction 
by  Leonhard  Euler  (1707-1783),  born  in  Switzerland  of  Dutch  descent,  in  a 
work  published  in  1748.  They  were  simultaneously  introduced  in  England 
by  Thomas  Simpson  (1710-1761),  professor  at  Woolwich,  in  his  Trigonom- 
etry, published  in  1748.  [See  Ball,  A  Short  History  of  Mathematics,  pp. 
215,  367.]  When  first  used  these  names  referred,  not  to  certain  ratios  con- 
nected with  an  angle,  but  to  certain  lines  connected  with  circular  arcs  sub- 
tended by  the  angle.  This  is  explained  in  Art.  79,  which  the  student  can 
easily  read  at  this  time.     See  Art.  80,  Notes  2,  3. 


12.1 


THE   TRIGONOMETRIC  BATIOS, 


23 


^h. 


sill  A  =  opposite  side,  " 
hypotenuse, 

.      adiacent  side, 

cos  A  =  —^ i 

hypotenuse, 


.^-——        ^  tan  A  =  opposite  side, 
V  /{  adjacent^ side, 

^  cot  A  -  ^^3^^^^^  side, 
opposite  side, 


see^=  hypotenuse, 
adjacent  side, 

cosec^=  t»yP°te""^e, 
opposite  side, 

[The  word  perpendicular  is  sometimes  used  instead  of  opposite  sidCy  and 
base  instead  of  adjacent  side.^ 

It  is  necessary  that  these  definitions  be  thoroughly  memorized. 


rx 


EXAMPLES. 

N.B.  The  student  is  requested  to  preserve  the  work  and  results  of  these 
Exs.  for  purposes  of  future  reference. 

1.  In  AMP  (Fig.  2)  give  the  trigonometric  ratios  of  angle  APM.  Note 
what  ratios  of  angles  A  and  P  are  equal. 

2.  In  Figs.  45  a,  45  b,  Art.  46,  give  the  trigonometric  ratios  of  the  various 
acute  angles. 

3.  Find  the  trigonometric  ratios  of  the  acute  angles  in  the  triangles  in  Exs. 
8-10,  Art.  8 ;  Exs.  11-13,  Art.  9;  Exs.  2-6,  Art.  11. 

4.  In  a  triangle  PQB  right-angled  at  §,  the  hypotenuse  PB  is  10  in. 
long,  and  the  side  QB  is  7.  ]Find  the  trigonometric  ratios  of  the  angles  P 
and  B.     Note  what  ratios  of  P  and  B  are  equal. 

5.  For  each  of  the  angles  in  Ex.  4,  and  for  each  of  any  three  of  the  angles 
in  Ex.  3,  calculate  the  following,  and  make  a  note  of  the  result.  [Let  x  de- 
note the  angle  whose  ratios  are  being  considered.] 

(1)  sin  ic  cosec  x  (2)  cos  x  sec  x  (3)  tan  x  cot  x 

(4)  sin2  X  +  cos2  x  (5)  sec^  x  —  tan2  x  (6)  cosec^  x  -  cot^  x 


(7)  tana;  - 


sinx 
cosx 


(8)  cot  X  — 


sin  X 


6.    Make  the  same  calculations  for  angle  A  in  Fig,  2,  Art.  12. 


24 


PLANE  TRIGONOMEfUT. 


[Ch.  II. 


13.  Definite  and  invariable  connection  between  (acute)  angles 
and  trigonometric  ratios.  It  is  important  that  the  following  prin- 
ciples be  clearly  understood: 

(1)  To  each  value  of  an  angle  there  corresponds  hut  one  value  of 
each  trigo7iometric  ratio. 

(2)  Tujo  unequal  acute  angles  have  different  trigonometric  ratios. 

(3)  To  each  value  of  a  trigonometric  ratio  there  corresponds  hut 
one  value  of  an  acute  angle. 

(1)  In  Fig.  2,  Art.  12,  from  any  point  ^S'  in  AB  draw  ST  per- 
pendicular to  AL.  Let  angle  B  (Fig.  3)  be  equal  to  A,  and  from 
any  point  G  in  one  of  the  lines  containing  angle  B  draw  GK 
perpendicular  to  the  other  line.    Then,  by  definition  (3),  Art.  12, 


sin  A  (in  AMP)  =  ^, 


^inAimAST) 


ST 
AS' 


sm  B  (=  sm  ^)  =  — . 

But  the  triangles  AMP,  AST,  BKG,  are  mutually  equi- 
angular. Hence  the  sides  about  the  equal  angles  are  propor- 
tional, and 

-  MP^ST^KG 
AP      AS      BG' 

Therefore  all  angles  equal  to  A  have  the  same  sine.     In  like 
manner,  these  angles  can  be  shown  to  have  the  same  tangent, 
secant,  etc.* 

(2)  Let  EAL  and  EALi  be  any  two  un- 
equal acute  angles,  placed,  for  convenience, 
so  as  to  have  a  common  vertex  A  and  a 
common  boundary  line  AR.  From  any 
point  P  on  AL  draw  PM  perpendicular 
^  to  AR.  Take  AP^  =  AP,  and  draw  P,M, 
perpendicular  to  AR.     Then 


*  In  Euclid's  text  on  geometry,  the  properties  of  similar  triangles  are  con- 
sidered in  Bk.  VI.  Pupils  who  study  Euclid  and  have  not  reached  Bk.  VI. 
can  be  helped  to  understand  these  properties  by  means  of  a  few  exercises 
like  those  referred  to  in  Ex.  3,  Art.  12. 


TABLES.  25 


sm  HAL  =  -—-,    sill  RAL^  =     ^   ^- 


But  M^P,  >  MP,  and  ^Pj  =  .4P; 

hence  sin  RALi  >  sin  RAL. 

In  a  similar  manner  the  other  ratios  can  be  shown  to  be  respec- 
tively unequal. 

Ex.  In  this  construction  AP\  is  taken  equal  to  AP.  Why  does  this  not 
affect  the  generality  of  the  proof  ? 

(3)  This  property  follows  as  a  corollary  from  (1)  and  (2). 

The  trigonometric  ratios  for  angles  from  0°  to  90°  are  arranged 
in  tables.  In  some  tables  the  calculations  are  given  to  four  places 
of  decimals,  in  others  to  five,  six,  or  seven  places.  There  are  also 
tables  of  the  logarithms  of  the  ratios  (or  of  the  logaritlims  in- 
creased by  10),*  which  vary  in  the  number  of  places  of  decimals 
to  which  the  calculations  are  carried  oufc.  The  student  is  advised 
to  examine  a  table  of  the  trigonometric  ratios  at  this  time.  A 
good  exercise  will  consist  in  finding  the  logarithms  of  some  of  the 
sines,  tangents,  etc.,  adding  10  to  each  logarithm,  and  comparing 
the  result  with  that  given  in  the  table  of  Logarithmic  sines,  tan- 
gents, etc.  [What  are  denoted  as  Natural  sines  and  cosines  in 
the  tables,  are  merely  the  actual  sines  and  cosines,  which  have  been 
discussed  above ;  the  so-called  Logarithmic  sines  and  cosines  are 
the  logarithms  of  the  Natural  sines  and  cosines  with  10  added.] 
A  book  of  logarithms  and  trigonometric  ratios  is  the  principal  help 
and  tool  in  solving  most  of  the  problems  in  practical  trigonom- 
etry; and  hence,  proficiency  in  using  the  tables  is  absolutely 
necessary.  The  larger  part  of  the  numerical  answers  in  this  book 
have  been  obtained  with  the  aid  of  a  five-place  table.  Those  who 
use  six-place  or  seven-place  tables  will  reach  more  accurate  results. 

EXAMPLES. 

1.  Compare  each  of  the  ratios  of  BALi  with  the  corresponding  ratio  of 
BAL. 

2.  Suppose  that  the  line  AB  (Fig.  4)  revolves  about  ^  in  a  counter-clock- 
wise direction,  starting  from  the  position  AM:  show  that,  as  the  angle  MAL 

*  These  are  usually  called  Logarithmic  sines,  tangents,  etc. 


26  PLANE  TRIGONOMETRY.  [Ch.  U. 

11 

increases,  its  sine,  tangent,  and  secant  increase,  and  its  cosine,  cotangent™' 
and  cosecant  decrease.     Test  this  conclusion  by  an  inspection  of  a  table  of 
Natural  ratios. 

3.  Find  by  tables,  sin  17°  40',  sin  43°  25'  10",  sin  76°  43',  sin  83°  20'  25", 
cos  18°  10',  cos  37°  40'  20",  cos  61°  37',  cos  72°  40'  30",  tan  37°  40'  20", 
tan  79°  37'  30",  cot  42°  30',  cot  72°  25'  30".  Log  sin  37°  20',  Log  sin  70°  21'  30", 
Log  cos  30°  20'  20",  Log  cos  71°  25',  Log  tan  79°  30'  20",  Log  cot  48°  20'  40". 

4.  Find  the  angles  corresponding  to  the  following  Natural  and  Logarithmic 
ratios : 

sine  =  .15327,  sine  =  .62175,  sine  =  .82462,  sine  =  .84316, 

cosine  =  .85970,        cosine  =  .61497,       cosine  =  .84065,       cosine  =  .60165. 

tangent  =  .42482,     tangent  =  .60980,    tangent  =  1.6820,     tangent  =  2.4927, 

Log  sine  =  9.79230,  Log  sine  =  9.94215, 

Log  cosine  =  9.96611,         Log  cosine  =  9.74743, 
Log  tangent  =  9.82120,      Log  tangent  =  10.37340. 

14.  Practical  problems.  The  problems  in  this  article  are  in- 
tended to  help  the  learner  to  realize  more  clearly  and  strongly  the 
meaning  and  the  usefulness  of  the  ratios  which  have  been  defined 
in  Art.  12.  The  student  is  earnestly  recommended  to  try  to  solve 
the  first  three  problems  below  witliout  help  from  the  book.  He 
will  find  this  to  be  an  advantage,  whether  he  can  solve  the  prob- 
lems or  not.  If  he  can  solve  them,  then  he  will  have  the  pleas- 
urable feeling  that  he  is  to  some  extent  independent  of  the  book ; 
and  he  will  thus  be  encouraged  and  strengthened  for  future  work. 
Should  he  fail  to  solve  them,  he  will  have  the  advantage  of  a 
closer  acquaintance  with  the  difficulties  in  the  problems,  and  so  • 
will  observe  more  keenly  how  these  difficulties  are  avoided  or 
overcome.  Throughout  this  course  the  student  will  find  it  to  be 
of  immense  advantage  if  he  will  think  and  study  over  the  subject- 
matter  indicated  in  the  headings  of  the  articles  and  make  some 
kind  of  an  attack  on  the  problems  before  appealing  to  the  book  for 
help.  If  he  follows  this  plan,  his  progress,  in  the  long  run,  will 
be  easier  and  more  rapid,  and  his  mental  power  more  greatly 
improved  than  if  he  is  content  merely  to  follow  after,  or  be  led 
by,  the  teacher. or  author. 


14.] 


PUACncAL  PROBLEMS. 


2T 


EXAMPLES.      " 

1.    Construct  the  acute  angle  whose  cosine  is  |.     What  are  its  other  trigo- 
nometric ratios  ?     Find  the  number  of  degrees  in  the  angle. 

The  definition  of  the  cosine  of  an  angle  shows  that  the  required  angle  is 
equal  to  an  angle  in  a  certain  right-angled  triangle,  namely,  the  triangle  in 
which  "  the  side  adjacent  to  the  angle  is  to  the  hypote- 
nuse in  the  ratio  2:3."  Thus  the  lengths  of  this  side 
and  hypotenuse  can  be  taken  as  2  and  3,  6  and  9,  200 
and  300,  and  so  on.  Taking  the  lengths  2,  3,  (these 
numbers  being  simpler  and,  accordingly,  more  conven- 
ient than  the  others),  construct  a  right-angled  triangle 
AST  which  has  side  AS  =  2,  and  hypotenuse  AT  =Z. 
The  angle  A  is  the  angle  required,  for  cos  A  =  ^. 

V5 


Now 


/S'T  =  V32-22 


Hence,  the  other  ratios  are 

sin  ^  =  ^  =  .7454,  tan  ^  =  ^ 
3  '  2 


sec  J.  =  -  =  1.5000,  cosec^ 


2.2361. 


1.1180,  cot  ^  =  -^  =  .8944, 
V5 
3 


1.3416. 


2  '  V5 

The  measure  of  the  angle  can  be  found  in  either  one  of  two  ways,  viz. :  (a) 

by  measuring  the  angle  with  the  protractor  ;  (6)  by  finding  in  the  table  the 

angle  whose  cosine  is  |  or  .6667.     The  latter  method  shows  that  A  =  48°  11'  22". 

[Compare  the  result  obtained  by  method  (a)  with  the  value  given  by  method 

2.   A  right-angled  triangle  has  an  angle  whose  cosine  is  f ,  and  the  length 
^    of  the  hypotenuse  is  50  ft.     Find  the  angles  and  the 
lengths  of  the  two  sides. 

By  method  shown  in  Ex.  1,  construct  an  angle  A 
whose  cosine  is  |.  On  one  boundary  line  of  the  angle 
take  a  length  AG  to  represent  50  ft.  Draw  GK  per- 
pendicular to  the  other  boundary  line. 


Fig.  6. 


cos^ 


AK 
AG 


Cos^  =  f  =  .6666  ..., 

.-.  ^  =  48°11'22^ 

.• 

-B  =  90  -  ^  =  41°  48'  38", 

=  .6666-., 

sin^=^^ 
3 

=  50  X  .6666  . 

.    KG      V5 
*'       "  AG~    3\ 

=  33.333  ..., 

.-.  /iG^  =  :^  X  50  =  37.27 

(Ex.  1) 


28 


PLANE  TRIGONOMETRY. 


[Ch.  II. 


•  The  problem  may  also  he  solved  graphically  as  follows.     Measure  angles 
A,  G,  with  the  protractor.     Measure  AK^  KG  directly  in  the  figure. 

3,  A  ladder  24  ft.  long  is  leaning  against  the  side  of  a 
building,  and  the  foot  of  the  ladder  is  distant  8  ft.  from 
the  building  in  a  horizontal  direction.  What  angle  does  the 
ladder  make  with  the  wall  ?  How  far  is  the  end  of  the 
ladder  from  the  ground  ? 

Graphical  method.  Let  AC  represent  the  ladder,  and 
BC  the  wall.  Draw  AC^  AB,  to  scale,  to  represent  24  ft. 
and  8  ft.  respectively.  Measure  angle  ACB  with  the  pro- 
tractor.    Measure  BC  directly  in  the  figure. 

Method  of  computation. 


BC  =  ^AG^  -  AB^  =  V676  -  64  =  V6l2  =  22.63  ft. 


sin  ^05  =  ^  =  A  =  .33333, 

.-.  ^0J5  =  19°  28' 16". 

4.  Find  tan  40°  by  construction  and  measurement. 
With  the  protractor  lay  off  an  angle  SAT  equal  to 
40°.  From  any  point  P  in  J.  T  draw  PB  perpendicu- 
larly to  AS.     Then  measure  AB.,  RP,  and  substitute 

RP 


the  values  in  the  ratio,  tan  40°  = 


AR 


Compare  the 


result  thus  obtained  with  the  value  given  for  tan  40°  in 
the  tables.* 

5.  Construct  the  angle  whose  tangent  is  f .  Find  its  other  ratios.  Meas- 
ure the  angle  approximately,  and  compare  the  result  with  that  given  in  the 
tables.  Draw  a  number  of  right-angled,  obtuse-angled,  and  acute-angled 
triangles,  each  of  which  has  an  angle  equal  to  this  angle. 

6.  Similarly  for  the  angle  whose  sine  is  | ;  and  for  the  angle  whose  co- 
tangent is  3. 

7.  Similarly  for  the  angle  whose  secant  is  2^  ;  and  for  the  angle  whose 
cosecant  is  3^. 

8.  Find  by  measurement  of  line«  the  approximate  values  of  the  trigo- 
nometric ratios  of  30°,  40°,  45°,  50°,  55°,  60°,  70° ;  compare  the  results  with 
the  values  given  in  the  tables. 


*  The  values  of  the  ratios  are  calculated  by  an  algebraic  method,  and  can 
be  found  to  any  degree  of  accuracy  that  may  be  required. 


15.] 


thigonometric  hatios  of  45°. 


29 


If  any  of  the  folloiving  constructions  asked  for  is  impossible,  explain  why 
it  is  so. 

9.  Construct  the  acute  angles  in  the  following  cases  :  (a)  When  the  sines 
are  ^,  2,  ^,  f ;  (&)  when  the  cosines  are  |,  ^,  3,  .3;  (c)  when  the  tangents 
are  3,  4,  |,  ^ ;  (d)  when  the  cotangents  are  4,  2,  |,  .7  ;  (e)  when  the  secants 
are  2,  3,  |-,  |,  4 J  ;  (/)  when  the  cosecants  are  3,  2.5,  .4,  f. 

10.  Find  the  other  trigonometric  ratios  of  the  angles  in  Ex.  9.  Find  the 
measures  of  these  angles,  (a)  with  the  protractor,  (&)  by  means  of  the  tables. 

11.  What  are  the  other  trigonometric  ratios  of  the  angles :  (1)  whose  sine 

is-;   (2)  wnose  cosine  is -;   (3)  whose  tangent  is -;    (4)   whose  cotangent 
b  b  b  ■ 

is-;   (5)  whose  secant  is -;    (6)  whose  cosecant  is  -  ? 
b    ^  ^  &     ^  ^  b 

12.  A  ladder  .32  ft.  long  is  leaning  against  a  house,  and  reaches  to  a  point 
24  ft.  from  the  ground.     Find  the  angle  between  the  ladder  and  the  wall. 

13.  A  man  whose  eye  is  5  ft.  8  in.  from  the  ground  is  on  a  level  with,  and 
120  ft.  distant  from,  the  foot  of  a  flag  pole  45  ft.  8  in.  high.  What  angle  does 
the  direction  of  his  gaze,  when  he  is  looking  at  the  top  of  the  pole,  make  with 
a  horizontal  line  from  his  eye  to  the  pole  ? 

14.  Find  the  ratios  of  45°,  60°,  30°,  0°,  90°,  before  reading  the  next  article. 


15.  Trigonometric  ratios  of  45°,  60°,  30°,  0°,  90°.     The  ratios  of 
certain  angles  which  are  often  met  will  now  be  found. 

A.   Ratios  of  45°.     Let  AMP 

be  an  isosceles  right-angled  tri- 
angle, and  let  each  of  the  sides 
about  the  right  angle  be  equal 
to  a. 

The  angle 

^  =  45°,  and^P=aV2. 


*.  sin  45°  =  sin  ^  = 


MP 


Fig.  10. 


^P      aV2     V2 
By  using  the  same  figure  it  can  be  shown  that 


cos  45°  =- 


V2' 
sec  45° 


tan  45°  =  1,  cot  45°  =  1, 

V2,  cosec  45°  =  V2. 


so 


PLANE  TlitGONOMETMV. 


[Ch.  It 


The  sides  of  triangle  AMF  are  proportional  to  1,  1,  V2. 
Hence,  in  order  to  produce  the  ratios  of  45°  quickly,  it  is  merely 
necessary  to  draw  Yig.  10 ;  from  this  figure  the  ratios  of  45°  can 
^B  J,         be  read  off  at  once. 

B,  Ratios  of  30°  and  60°, 
Let  ABC  be  an  equilateral 
^  triangle.  From  any  vertex  B 
draw  a  perpendicular  BD 
to  the  opposite  side  AC. 
Then  angle  DAB  =  60°,  an- 
gle ABD  =  30°. 


A 

[\ 

/wt° 

V 

ovrN 

\ 

/ 

A\ 

A 

/C\ 

4^        a 

^D 

G 

1 

Fig 

11. 

Fig,  12. 

liAB  =  2a,  then  AD  =  a,  and  DB  =-V4.a^-d'  =  aVS. 

.-.  sin  60°  =  sin  DAB  =  ^  =  ^  =  2^, 
AB       2a        2 


By  using  the  same  figure  it  can  be  shown  that 

cos  60°  =  1,  tan  60°  =  V3,  cot  60°        "'■ 

2 


V3' 


sec  60°  =  2. 


cosec  60°  = 


Also, 
Similarly, 

cos  30° 


sin  30°  =  sin  ABD  =  — 


AB      2  a 


V3 


V3 


tan  30' 


1 

V3' 


sec  30°  = 


V3' 


cot  30°  =  V3, 
cosec  30°  =  2. 


In  ADB  the  sides  opposite  to  the  angles  30°,  60°,  90°,  are 
respectively  proportional  to  1,  V3,  2.  Hence,  in  order  to  produce 
the  ratios  of  30°,  60°,  at  a  moment's  notice,  it  is  merely  necessary 
to  draw  Fig.  12,  from  which  these  ratios  can  be  imjuediately  read 
off. 


C.   Ratios  of  0"  and  90° 

read  now. 


The  algebraical  note,  Art  76,  may  he 


15.] 


TBIGONOMETRIC  RATIOS   OF  0°   AND  90°, 


31 


Let  the  hypotenuse  in  each  of  the  right-angled  triangles  in 
Fig.  13  be  equal  to  a. 

MP 


Bin  MAP  = 


cos  MAP  = 


AP' 
AM 
AP' 


It  is  apparent  from  this  figure  that  if 
the   angle  MAP  approaches    zero,  then 
the  perpendicular  MP  approaches  zero,     ^ 
and  the  hypotenuse  AP  approaches  to  an 
equality  with  AM;  so  that,  finally,  if  MAP  =  0,  then  MP  =  0, 
and  AP  =  AM.     Therefore,  when  MAP  —  0,  it  follows  that : 


sin  0°  =  -  =  0, 
a 


cos  0°  =  -  =  1, 


tan  0°  =  -  =  0, 
a 


cot  0°  =  ^ 


00. 


sec  0°  =  - 
a 


1, 


cosec  0°  =  -  =  00 


As  MAP  approaches  90°,  AM  approaches  zero,  and  MP  ap- 
proaches to  an  equality  with  AP.  Therefore,  when  MAP  =  90°, 
it  follows  that : 


sin  90°  =  -  =  1, 
a 


cos  90°  =  -  =  0, 
a 


tan  90" 


0 


=  00, 


sec  90°  =  ^=00, 


cot  90°  =  5  =  0,      cosec  90°  =  -  =  1. 
a  a 


EXAMPLES. 

N,  B.    Read  the  first  few  lines  of  Art.  17  before  attacking  the  problems. 
Find  the  numerical  value  of 


2.   sec2  30°  +  tan3  45°. 
4.   cos  0°  sin  45°  +  sin  90°  sec2  30°. 
6.    3  tan3  30°  sec^  60°  sin2  90°  tan2  45°. 
8.   2  sin^  30°  tan^  60°  cos^  0°. 
9.    X  cot8  45°  sec2  60°  =  11  sin2  90°  ;  find  x. 
10.   x(cos  30°  +  2  sin  90°  +  3  cos  45°  -  sin2  60°)  =  2  sec  0°  -  6  sin  90° ;  find  x. 


1.  sin  60°  +  2  cos  45°. 

3.  sinS  60°  +  cot8  30°. 

5.  4  cos2  30°  sin2  60°  cos2  0°. 

7.  10  cos*  45°  sec6  30°. 


-L-^ 


b2  PLANE  TRIGONOMETRY,  [Ch.  II. 

16.  Relations  between  the  trigonometric  ratios  of  an  acute  angle 
and  those  of  its  complement.  When  two  angles  added  together 
make  a  right  angle,  the  two  angles  are  said  to  be  complementary ^ 
and  each  angle  is  called  the  complement  of  the  other. 

For  example,  the  acute  angles  in  a  right-angled  triangle  are  complemen- 
tary ;  the  complement  of  ^  is  90°  —  ^  ;  the  complement  of  27°  is  63°. 

Ex.  1.    What  are  the  complements  of  10°,  12°  30',  47°,  56°  27',  35°  ? 
Ex.  2.    What  angles  are  complementary  to  23°,  42°,  51°,  78°,  86°  ? 

In  Fig.  2,  Art.  12,  the  angle  APM  is  the  complement  of  the 
angle  A.    Now, 

sinP  =  4|;    tanP  =  ^,        secP  =  ^, 
AP'  MP'  MP' 

cosP  =  ^,     cotP  =  i^,    cosecP  =  4^. 
AP  AM  AM 

Comparison  of  these  ratios  with  the  ratios  of  A  in  (2),  Art.  12, 
shows  that 

cos  A  =  sin  P,     tan  A  =  cot  P,         sec  A  =  cosec  P, 

sin  A  =  cos  P,      cot  A  —  tan  P,     cosec  A  =  sec  P. 

These  six  relations  can  be  expressed  briefly : 

Each  trigonometric  ratio  of  an  angle  is  eqnal  to  the  corresponding 
co-ratio  of  its  complement. 

Ex.    Compare  the  ratios  of  30°  and  60°  ;  of  0°  and  90°. 

17.  Exponents  in  trigonometry. 

When  a  trigonometric  ratio  has  an  exponent,  a  particular  way  of  placing 
the  exponent  has  been  adopted.     For  example, 

(since) 2  is  written  sin^jc. 

There  is  no  ambiguity  in  the  second  form,  and  the  advantage  is  apparent. 

Thus,  cosmic,  tan^a;,  sec* a;,  represent  or  mean  (cosec) 3,  (tanx)^  (seccc)*. 
There  is  one  exponent,  however,  which  must  not  be  written  with  the 
brackets  removed.     This  exception  is  the  exponent  —  1.     Thus,  for  example, 

(cos  ic)-i,  which  means ,  must  never  be  written  cos-i  x.     The  reason 

cosx 
for  this  is  that  the  symbol  cos-i  x  is  used  to  represent  something  else.     This 
symbol  denotes  the  angle  whose  cosine  is  x,  and  is  read  thus,  or  is  read  "  the 


16-18.]  BELATIONS  BETWEEN  RATIOS.  33 

anti-cosine  of  x,"  '.'the  inverse  cosine  of  x,"  "cosine  minus  one  x."  The 
number  —  1 ,  which  appears  in  cos-i  x,  is  not  an  exponent  at  all,  but  is  merely 
part  of  a  symbol. 

Suppose  that  (a)  "  the  sine  of  the  angle  J.  is  f." 

The  latter  idea  can  also  be  expressed  by  saying 

(&)  "J.  is  the  angle  whose  sine  is  f  "  ; 

or,  more  briefly,  by  saying, 

(c)  "  ^  is  the  anti-sine  of  f." 

The  two  ways,  (a),  (c),  of  expressing  the  same  idea  can  be  indicated  still 
more  briefly  by  equations,  viz., 

sin  ^  =  f ,     A  =  sin-i  f . 

Thus,  (sinic)-i  and  sin-^a;  mean  very  different  things;  for  (sincc)-i  is 

,  which  is  a  number,  and  sin-i  x  is  an  angle. 

sinic 

Note.     The  symbols  sin-ia;,  cos~ia;,  •••,  are  considered  in  Art.  88. 

Ex.    Express  sin  -4  =  |,  cos  jc  =  |,  tan  C  =  4,  sec  ^  =  9,  cosec  A  =  ^,m 

the  inverse  form. 

18.  Relations  between  the  trigonometric  ratios  of  an  acute  angle. 

[N.B.  Some  relations  between  these  ratios  may  have  been  noticed  or 
discovered  by  the  student  in  the  course  of  his  preceding  work.  If  so,  they 
should  now  be  collected,  so  that  they  can  be  compared  with  the  relations 
shown  in  this  article.  ] 

Some  of  the  preceding  exercises  have  shown  that  when  one 
trigonometcic  ratio  of  an  angle  is  known,  the  remaining  Jive 
ratios  can  be  easily  determined.  This  at  least  suggests  that  the 
ratios  are  related  to  one  another.  In  what  follows,  A  denotes 
any  acute  angle. 

A.   Reciprocal  relations  between  the  ratios. 
Inspection  of  the  definitions  (3),  Art.  12,  shows  that : 

(a)  sin  A  = -,  cosec  A  =  -: -,  or,  sin  A  cosec  A  =  l; 

cosec  A  sm  A 


(b)  cos  A  = -,  sec  A  = -,  or,      cos  ^  sec  ^  =  1 ; 

sec  A  cos  A 

(c)  tan^  = -,  cotJ.  = 7j  or,     tan-4cot^  =  l. 

cot  -4  tan  4 


(1) 


34  PLANE  TBIGONOMETBT.  [Ch.  II. 

B.    The  tangent  and  cotangent  in  terms  of  the  sine  and  cosine. 
In  the  triangle  AMP  (Fig.  2,  Art.  12), 

MP  AM 

^^^  *  _  MP  _AP      unA^    _^  ^  _  AM  _  ^lP_cos^     (9\  m\ 

^P  ^P 

O.   Relations  between  the  squares  of  certain  ratios. 

In  the  triangle  AMP  (Fig.  2,  Art.  12),  indicating  by  mP  the 
square  of  the  length  of  MPj 

MP^  +  AM^  =  AP, 

On  dividing  each  member  of  this  equation  by  -<flP  ,  AM ,  MP , 
in  turn,  there  is  obtained 

fMpy,rAMy_fApy 

\AP)\~APJ~\AP)' 

(MP\    fAMV^  fALY 
\AMJ      \AmJ      \AmJ' 

fMPy    fAMY^  fAP\^ 
\MP)      \MPJ      \MP)' 

In  reference  to  the  angle  A,  these  equations  can  be  written : 

sin^  A  +  cos^  A  =  l, 

tan^^  +  l  =sec2^, (4) 

1  +  cot^  A  =  cosec^  A. . 

Note  1.  The  relations  shown  above  are  true,  not  only  for  acute  angles, 
but  for  all  angles.     This  is  shown  in  Art.  44. 

Note  2.   Relations  (1)  have  a  practical  bearing  on  the  construction  and 

the  use  of  tables.    Thus,  for  example,  since  cos  A  = ,  a  table  of  natural 

sec  A 
cosines  can  be  transformed  into  a  table  of   natural    secants   by  merely 
taking  the  reciprocals  of  the  cosines.     Again,  in  logarithmic  computation, 

since  sec  A  = ,  log  sec  ^  =  —  log  cos  A. 

cos  A 


18.]  EXAMPLES.  35 

Note  3.  An  equation  involving  trigonometric  ratios  is  a  trigonometric 
equation.  Thus,  for  example,  tan  A=l.  One  angle  which  satisfies  this 
equation  is  the  acute  angle  A  =  45".  Other  solutions  can  be  found  after 
Arts.  84-87  have  been  taken  up. 

EXAMPLES. 

A  few  simple  exercises  are  given  below,  the  solution  of  which  brings  in 
the  relations  shown  in  this  article.  These  exercises  are  algebraic  in  char- 
acter ;  collections  of  exercises  of  this  kind  are  given  also  in  other  places  in 
this  book.  In  the  following  examples,  the  positive  values  of  the  radicals  are 
to  be  taken.     The  meaning  of  the  negative  values  is  shown  in  Art.  44. 

1.    Given  that  sin  A  =  ^,  find  the  other  trigonometric  ratios  of  A  by- 
means  of  the  relations  shown  in  this  article. 

cosec  A  =  -J—  =  2  ;  cos  ^  =  Vl  -  sin2  A=^;  sec  A-     ^     -   ^   ■ 


sin^         '  '"  2   '  cos  A      ^3' 

tan^=^HL^  =  -l^;    cot^  =  -l-=V3. 
cos  A     Vo  t^^  ^ 

These  results  may  be  verified  by  the  method  used  in  solving  Exs.  1,  5-7, 
Art.  14. 

2.   Express  all  the  ratios  of  angle  A  in  terms  of  sin  A. 

sin  A  sinA, 


sm 


^  =  sin  ^  ;    cos  J.  =  Vl  —  sin^  A ;    tan  A 


cos  A     Vl-sin2^ 


ootA  =  -±-  =  ^^^^^^;  secA=-^=  ^  ;  cosec^=-J-. 

tan^l  sin  J.  cos  J.     Vl  — sin'-^J.  smA 

3.  Prove  that        :; -. — -  +  - -. — -  =  2sec2A 

1  —  sm  J.     1  +  sm  J. 

1         + 1 = ? :=_2_  =  2sec2A 


1  —  sin  Jl     1  +  sin  J.     1  —  sin2  A     cos2  A 

4.  Prove  that  sec*  ^-1  =  2  tan2  A  +  tan*  A. 

sec*  ^  -  1  =  (sec2  ^)2  _  1  =  (1  +  tan2  ^)2  _  1  =  2  tan2  A  +  tan*  A, 

5.  Solve  the  equation  4  sin  ^  —  3  cosec  ^  =  0. 

4sin0 ^=0. 

sin  d 

,;  4sin2^-3  =  0, 
.-.  sin2  e  =  -,    .'.  sin  6=  +  ^,  and  sin  6=  — -^• 


36  PLANE  TRIGONOMETRY.  [Ch.  11. 

On  taking  the  plus  sign,  one  solution  is  the  acute  angle  d  =  60° ;  other 
solutions  will  be  found  later.  For  the  minus  sign  there  is  also  a  set  of  solu- 
tions ;  these  will  be  found  later. 

6.  Solve  2  sin2  6  cosec  6  —  6  +  2  cosec  ^  =  0, 

sin  6  sin  6 

2  sin2  ^  -  5  sin  ^  +  2  =  0, 

(2  sin  ^  -  1)  (sin  6-2)  =0. 

.-.  sin  ^  =  I,  and  sin ^  =  2. 

The  acute  angle  whose  sine  is  |  is  30°  ;  hence  6  =  30°  is  one  solution.  The 
sine  cannot  exceed  unity  ;  hence  sin  ^  =  2  does  not  afford  any  solution. 

7.  Given  cos  J.=f,  sin  J5=|,  tan  0=2,  cotZ)=f,  secJS'=3,  cosec  1^=2.5  ; 
find  the  other  trigonometric  ratios  of  A,  B,  O,  i>,  E,  F,  by  the  algebraic 
method.     Verify  the  results  by  the  method  used  in  Art.  14. 

8.  Find  by  the  algebraic  method  the  ratios  required  in  Exs.  1,  5-7,  10, 
11,  Art.  14. 

9.  Express  all  the  trigonometric  ratios  of  an  angle  A  in  terms  of : 
(a)  cos  A  ;  (&)  tan  A  ;  (c)  cot  A  ;  (d)  sec  A  ;  (e)  cosec  A.  Arrange  the 
results  and  those  of  Ex.  2,  neatly  in  tabular  form. 

Prove  the  following  identities : 

10.  (sec2  ^  -  1)  cot2  ^  =  1 ;  cos  y1  tan  yl  =  sin  ^  ;  (1  -  sin^  A)  sec^  A  =  l. 

11.  sin2  6  sec2  6  =  sec^  ^  -  1 ;  tan2  6  -  cot^  6  =  sec2  6  -  cosec^  6. 

io  1       I        1       _i.         sin  ^    I  cos  yl  _ .,  . 

sec2^     cosec2^  cosec  ^     sec^ 

(tan^  +  sec^)2  =  l±4^. 
^  ^       1-sin^ 

18.  sec2  A  +  cosec2  A  =  tan2  ^  +  cot2  ^  +  2 ; 

1  +  tan2  A  _  sin2  A .  cosec  A       _  ^^^  ^^ 

1  +  cot2  A     cos2^ '       cot  ^  +  tan  J. 

^^  cos^     ^     sin^     =sin^  +  cos^; 

1  —  tan  A     1  —  cot  ^ 

=  sec  J  +  tan  A  ;  sec*  A  —  sec2  A  =  tan*  ^  +  tan  2  A. 


sec  A  —  tan  A 


19.]  BUMMABY,  37 

Solve  the  following  equations  : 

15.   2  sin  e  =  2  -  cos  d.  16.  tan  ^  +  cot  ^  =  2. 

17.   tan  ^  +  3  cot  ^  =  4.  18.  6  sec2  0  -  13  sec  61  +  5  =  0. 

19.    8  sin2  ^  -  10  sin  ^  +  3  =  0.  20.  sin  ^  +  2  cos  6  =  2.2. 

19.  Summary.  In  this  chapter  important  additions  have  been 
made  to  the  knowledge  concerning  angles  that  one  gained  in 
geometry.  A  process  of  measuring  angles  has  been  introduced. 
The  close  connection  between  angles  and  the  ratios  of  lines  has 
been  emphasized.  It  has  been  shown  that  each  (acute)  angle 
has,  associated  with  it,  a  definite  set  of  six  numbers,  called 
trigonometric  ratios ;  and  it  has  been  seen  that  the  sets  of  num- 
bers are  different  for  different  angles.  It  has  also  been  shown 
that  the  seven  quantities  (namely,  the  angle  and  the  six  numbers) 
are  so  related,  that,  if  one  of  the  seven  be  given,  then  the  remaining 
six  can  he  detennined. 

A  few  applications  to  the  measurement  of  lines  and  angles  have 
been  made  in  some  of  the  preceding  articles.  The  next  two  chap- 
ters are  taken  up  with  a  formal  treatment  of  such  applications. 
It  should  be  stated,  however,  that  any  one  who  understands  the 
contents  of  this  chapter  is  in  possession  of  all  the  principles  which 
will  be  used  in  the  next  two  chapters,  and  can  proceed  directly  to 
the  solution  of  the  problems  given  there.  The  student  is  recom-") 
mended  to  attack  some  of  the  exercises  in  Chapters  III.,  IV.,  \ 
before  reading  the  explanations  given  in  the  text.  Attention  may 
again  be  given  to  the  first  part  of  Art.  14. 

N.B.  Questions  and  exercises  suitable  for  practice  and  review  on  the  sub- 
ject-matter of  Chapter  II.  will  be  found  at  pages  182,  183. 


CHAPTER   111. 

SOLUTION   OF   RIGHT-ANGLED  TRIANGLES. 

Before  the  solution  of  right-angled  triangles  is  entered  upon,  a 
few  remarks  will  be  made  on  the  solution  of  triangles  in  general. 
Some  of  the  ideas  expressed  in  Arts.  20-24  are  applicable  to  prac- 
tical problems  throughout  the  book. 

20.  Solution  of  a  triangle.  Every  triangle  has  three  sides  and 
three  angles.  These  six  quantities  are  called  the  parts  or  elements 
of  a  triangle.  Sometimes  one  or  several  of  the  parts  of  a  triangle 
are  known ;  for  instance,  the  three  sides,  two  angles  and  a  side, 
two  sides,  one  side,  three  angles,  and  so  on.  In  such  cases  the 
questions  arise:  Can  the  remaining  parts  be  found  or  deter- 
mined ?  and,  if  so,  by  what  method  shall  this  be  done  ?  The  pro- 
cess of  deducing  the  unknown  parts  of  a  triangle  from  the  known, 
is  called  solving  the  triangle,  or,  the  solution  of  the  triangle.  This 
Chapter  and  Chapter  VII.  are  concerned  with  showing,  in  detail, 
methods  of  solving  triangles.  There  are  two  methods  which  can 
be  used  to  find  (only  approximately,  in  general)  the  unknown 
parts  of  a  triangle  when  some  of  its  parts  are  given.     These 

methods  are: 

(a)  The  graphical  method; 

(6)  The  method  of  computation, 

21.  The  graphical  method.  This  method  consists  in  draiving  a 
triangle  which  has  angles  equal  to  the  given  angles,  and  sides 
proportional  to,  and  thus  representing  the  given  sides,  and  then 
measuring  the  remaining  parts  directly  from,  the  drawing. 

For  example,  a  triangle  has  two  sides  whose  lengths  are  10  ft.,  5  ft.,  and 
the  included  angle  is  28°  30' ;  the  third  side  and  the  other  angles  are  required. 

The  graphical  solution  is  as  follows :  Construct  a  triangle  QPR  having 
two  sides,  PQ,  PB,  representing  10  ft.,  5  ft.,  respectively,  on  some  con- 

88 


20-22.]  METHOD   OF  COMPUTATION.  39 

venient  scale,  and  with  their  included  angle,  QPB,  equal  to  28°  30',  as 
shown  in  Fig.  14.  Measure  the  angles  PBQ,  PQB  with  the  protractor; 
measure  the  side  PQ  and,  by  reference  to 
the  scale,  find  the  length  represented  by  PQ. 
[The  results  thus  obtained  may  be  compared 
with  those  obtainable  by  the  method  of  com- 
putation explained  in  Arts.  54,  57.  The  latter 
results  are  P  =  128°  26'  46",  Q  =  28°  3'  14", 
7^^  =  6.092.] 

The  conditions  necessary  and  sufficient  for  constructing  a  tri- 
angle, and  the  methods  of  drawing  triangles  that  satisfy  given  con- 
ditions, are  shown  in  plane  geometry  and  in  geometrical  drawing. 
It  is  obvious  that  the  grajjhical  method  can  be  employed  only  when 
the  values  of  the  parts  given  are  consistent  loith  one  another,  and 
when  the  parts  given  are  sufficient  in  number  to  determine  a  definite 
triangle.  For  instance,  suppose  that  one  is  asked  to  lind  the 
remaining  parts  of  a  triangle  one  of  whose  sides  is  10  inches 
long.  In  this  case  as  many  unequal  triangles  as  one  please,  can 
be  constructed,  all  of  which  will  satisfy  the  given  condition. 
Again,  a  given  side  and  a  given  angle  are  insufficient  data  on 
which  to  proceed  to  find  the  remaining  parts  of  a  triangle,  for 
there  is  an  infinite  number  of  unequal  triangles  which  can  have 
parts  equal  to  the  given  parts.  So  also  the  method  fails  if  three 
angles  be  given;  for  an  infinite  number  of  unequal  triangles  can 
be  drawn  whose  angles  are  equal  to  the  given  angles.  Again,  let 
it  be  required  to  find  the  angles  of  a  triangle  whose  sides  are 
10  feet,  40  feet,  GO  feet.  Such  a  triangle  is  impossible,  since  the 
length  of  one  side  (GO  feet)  is  greater  than  the  sum  of  the  lengths 
of  the  other  two.  One  more  instance  :  let  two  given  angles  be  85° 
and  105°  and  the  included  side  be  40  inches ;  this  triangle  is  im- 
l)ossible,  since  the  sum  of  the  two  given  angles  is  greater  than 
two  right  angles. 

22.  The  method  of  computation.  This  method  is  applicable  in 
precisely  the  same  cases  in  which  the  preceding  method  can  be 
employed ;  namely,  in  the  cases  in  which  the  parts  given  are  con- 
sistent with  one  another,  and  afford  conditions  sufficient  to  enable 
one  to  construct  a  definite  triangle.  This  will  be  fully  apparent 
later,  when  the  various  cases  will  be  treated  in  detail.     One  of 


40  PLANE  TlilGONOMETRY,  [Cii.  III. 

the  principal  purposes  of  this  book  is  to  show  the  different 
methods  of  computation  applicable  to  various  sets  of  given  condi- 
tions. Oyie  of  the  principal  objects  of  a  student  who  is  taking  a 
first  course  in  trigonometry  should  he  to  acquire  facility  and,  above 
all,  accuracy  in  using  these  methods  of  computation. 

23.  Comparison  between  the  graphical  method  and  the  method  of 
computation.  The  experience  gained  in  some  of  the  exercises  in 
the  preceding  chapter  has  probably  shown  the  student  that  he  can 
attain  much  greater  accuracy  by  using  the  method  of  computation 
than  by  using  the  graphical  method.  The  accuracy  of  the  results 
obtained  by  the  latter  method  depends  upon  the  carefulness  and 
skill  with  which  the  figures  are  drawn  and  measured;  in  the 
other  method,  accuracy  depends  upon  the  care  and  patience  em- 
ployed in  performing  arithmetical  work.  While  the  results 
attainable  by  the  gr.aphical  method,  even  in  the  case  of  skilled 
persons  with  excellent  drawing  instruments,  are  not  as  accurate 
as  the  results  attainable  by  the  other  method,  yet  they  are  often 
accurate  enough  for  practical  purposes.  When  the  computations 
required  are  long  and  complicated,  the  graphical  method  is  much 
the  more  rapid  of  the  two. 

There  are  several  reasons  why  it  is  advisable  for  the  learner  to 
use  the  graphical  method,  as  well  as  the  method  of  computation, 
in  solving  problems  in  this  course.  The  first  reason  is  that  the 
former  method  icill  serve  as  a  check  upon  the  latter.  With  or- 
dinary care  the  graphical  method  very  quickly  gives  a  fair 
approximation  to  the  result.  This  result  will  sometimes  show 
that  there  is  an  error  in  the  result  obtained  by  computation.  A 
little  error  in  arithmetic  may  yield  a  quantity  which  is  ten  times 
too  great  or  too  small ;  but  this  can  be  detected  at  once  if  the 
other  method  has  also  been  used.*  A  second  reason  for  using  the 
graphical  method  is  that  this  method  incidentally  provides  train- 
ing in  neat,  careful,  and  accurate  drawing ;  this  training  will  not 
only  be  a  benefit  in  itself,  but  will  be  of  very  great  advantage  in 
other  studies,  and  especially  in  the  applied  sciences.     A  third 


*  The  results  can  also  be  tested  by  methods  of  computation,  which  will  be 
sliown  in  due  course. 


23-25.]  GENERAL  BIEECTIONS.  41 

reason  is  that  the  pupil  will  gai7i  some  knowledge  and  experience 
of  a  method  that  is  used  in  other  subjects,  for  instance,  in  physics 
and  in  mechanics,  and  that  is  extensively  employed  by  engineers 
in  solving  problems  in  v/hich  the  computations  required  by  the 
other  method  may  be  overwhelmingly  cumbrous. 

24.  General  directions  for  solving  problems.  A  third  method  of 
approximating  to  the  magnitude  of  lines  and  angles  may  be  men- 
tioned here,  for  it  has  often  to  be  employed  in  practical  life.  In 
this  method  the  student  may  suppose  that  he  possesses  neither 
measuring  instruments,  drawing  materials,  nor  mathematical 
tables,  and  thereupon  he  may  give  an  off-hand  estimate  concerning 
the  magnitudes  required.  This  method  also  serves  as  a  check,  by 
showing  when  great  arithmetical  blunders  are  committed.  The 
pupil  is  advised  to  use  all  three  methods  in  working  each  practical 
problem  in  this  course,  and  to  do  so  in  the  following  order: 

(1)  Make  an  off-hand  estimate  as  to  what  the  magnitude  required 
may  he,  and  write  this  estimate  down; 

(2)  Solve  the  problem  by  the  graphical  method; 

(3)  Solve  the  problem  by  the  slower  but  more  accurate  and  reliable 
method  of  computation.  There  may  be  some  interest  found  in 
comparing  the  results  obtained  by  these  three  methods.  The 
exercise  in  judging  linear  and  angular  magnitudes  afforded  by 
the  first  method,  the  practice  in  neat  and  careful  drawing  neces- 
sary in  the  second,  and  the  training  in  accurate  computation 
given  by  the  third,  will  each  afford  some  benefit  to  the 
learner. 

25.  Solution  of  right-angled  triangles.     Let  ABC  be  a  right- 
angled  triangle,  C  being  the   right   angle.     In  ^b 
what  follows,  a,  b,  c,  denote  the  lengths  of  the 
sides  opposite  to  the  angles  A,  B,  C,  respectively. 
The  sides  and  angles  of  ABC  are  connected  by 
the  following  relations: 

„        „  '  r  .  .  (Geometry) 


42 


PLANE  TEIGONOMETBT. 


[Cu  IIL 


.    (Definitions  (3),  Art. 


(3),  (4)     sin^=-=cos5; 


(5),  (6)     cos^  =  -  =  sin^ 
c 

(7),  (8)     tan^  =  -=cot5; 
(9),  (10)  cot^  =  -  =  tan5. 


Note.    To  these  may  be  added :  sec  ^  =  -  =  cosec  B,  cosec  A  =  ~  =  secB. 

b  a 

These  relations,  however,  are  not  needed,  and  are  rarely  used  in  solving 
triangles,  for  they  are  equivalent  to  (3)-(6),  and  few  tables  give  secants  and 
cosecants.    See  Art.  18,  Note  2. 

Equation  (1)  shows  that  no  other  element  of  the  triangle  can 
be  derived  from  the  two  acute  angles  only.  Each  of  the  remain- 
ing equations,  (2)-(10),  involves  three  elements  of  the  triangle, 
and  at  least  two  of  these  elements  are  sides.  Hence,  in  order 
that  a  right-angled  triangle  be  solvable,  two  elements  must  be 
known  in  addition  to  the  right  angle,  and  one  of  these  must  be  a 
side.  If  any  two  of  the  elements  involved  in  equations  (2)-(10) 
are  known,  then  a  third  element  of  the  triangle  can  be  found 
therefrom.  Hence  the  following  general  rule  can  be  used  in 
solving  right-angled  triangles : 

When  in  addition  to  the  right  angle,  any  two  sides,  or  one  of  the 
acute  angles  and  any  07ie  of  the  sides,  of  a  right-angled  triangle  are 
Tcnown,  and'  another  element  is  required,  write  the  equation  involving 
the  required  element  and  two  of  the  known  elements,  and  solve  the 
equation  for  the  required  element. 

Eor  example,  suppose  that  a,  c  are  known,  and  that  A,  B,  b 
are  required.     In  this  case, 


sin.4  =  ^,     B 
c 


90 -A,    b'  =  Vc' 


The  quantities  A,  B,  b  can  be  found  from  these  equations.  If 
an  error  has  been  made  in  finding  A,  then  B  will  also  be  wrong. 
Hence  it  is  advisable  to  check  the  values  found  by  seeing  whether 
they  satisfy  relations   differing   from   those   already   employed. 


26,  27.]  CHECKS.      CASES.  43 

For  example,  check  formulas  which  may  be  used  in  this  case 
are: 

-  =  tan  B,    -=  cos  A. 
a  c 

26.  Checks  upon  the  accuracy  of  the  computation.  As  already 
pointed  out,  large  errors  can  be  detected  by  means  of  the  off-hand 
estimate  and  by  the  use  of  the  graphical  method.  The  calculated 
results  in  any  example  can  be  checked  or  tested  by  employing  rela- 
tions which  have  not  been  used  in  computing  the  results,  and 
examining  whether  the  newly  found  values  satisfy  these  relations. 
An  instance  has  been  given  in  the  preceding  article.  The  student 
is  advised  not  to  look  up  the  answers  until  after  he  has  tested  his 
results  in  this  way.  Verification  by  means  of  check  formulas  is 
necessary  in  cases  in  which  the  answers  are  not  given.  The  test- 
ing of  the  results  also  affords  practice  in  the  use  of  formulas  and 
in  computation.  When  a  check  formula  is  satisfied  it  is  highly 
probable,  but  not  absolutely  certain,  that  the  calculated  results 
are  correct. 

27.  Cases   in   the   solution   of    right-angled  triangles.     All   the 

possible  sets  of  two  elements  that  can  be  made  from  the  three 
sides  and  the  two  acute  angles  of  a  right-angled  triangle  are  the 
following : 

(1)  The  two  sides  about  the  right  angle. 

(2)  The  hypotenuse  and  one  of  the  sides  about  the  right  angle. 

(3)  The  hypotenuse  and  an  acute  angle. 

(4)  One  of  the  sides  about  the  right  angle,  and  an  acute  angle. 

(5)  The  two  acute  angles.  (This  case  has  already  been 
referred  to.) 

Some  examples  of  these  cases  are  solved.  The  general  method 
of  procedure,  after  making  an  off-hand  estimate  and  finding  an 
approximate  solution  by  the  graphical  method,  is  as  follows : 

First :  Write  all  the  relations  (or  formulas)  which  are  to  he  used 
in  solving  the  problem. 

Second:    Write  the  check  formulas. 

Third:  In  making  the  computations  arrange  the  ivork  as  neatly 
as  possible. 


44  PLANE  TRIGONOMETRY.  [Cu.  III. 

This  last  is  important,  because,  by  attention  to  this  rule,  the 
work  is  presented  clearly,  and  mistakes  are  less  likely  to  occur. 
The  computations  may  be  made  either  with  or  without  the  help 
of  logarithms.  The  calculations  can  generally  be  made  more 
easily  and  quickly  by  using  logarithms. 

Note  1.  Relations  (3),  (4),  Art.  25,  may  be  written:  a  =  csinj., 
a  =  c  cos  B.     'J'hese  relations  may  be  thus  expressed : 

A  side  of  a  right-angled  triangle  is  equal  to  the  product  of  the  hypotenuse 
and  the  cosine  of  the  angle  adjacent  to  the  side. 

A  side  of  a  right-angled  triangle  is  equal  to  the  product  of  the  hypotenuse 
and  the  sine  of  the  angle  opposite  to  the  side. 

Note  2.  Relations  (7),  (8),  Art.  25,  may  be  written:  a  =  &  tan  ^, 
a  =  &  cot  B.     These  relations  may  be  thus  expressed  : 

A  side  of  a  right-angled  triangle  is  equal  to  the  product  of  the  other  side 
and  the  tangent  of  the  angle  opposite  to  the  first  side. 

A  side  of  a  right-angled  triangle  is  equal  to  the  product  of  the  other  side 
and  the  cotangent  of  the  angle  adjacent  to  the  first  side. 

EXAMPLES. 

i.   In  the  triangle  ABC,  right-angled  at  C,  a  =  42  ft.,  6  =  56  ft.    Find 
B '   the  hypotenuse  and  the  acute  angles. 

I.    Computation  without  logarithms.    [Four-place 
tables.  ] 

tan  ^  =  ^  =  ^  =  .7500.  .-.  A  =  36°  62'. 2. 

b      56 
B=QO-A.  .♦.  ^  =  53°    7'.8. 

c  =  Va2  -f-  62  ^  V1764  +  3136.     .-.    c  =  70  ft. 
Check :        a=:ccosB  =  70  x  cos  53°  7'.8  =  70  x  .6000  =  42  ft. 

II.    Computation  with  logarithms. 
Given :  a  =  42  ft.  To  find  :*      A  = 

6  =  56  ft.  B  = 

c  = 
Formulas  :  tan  A  =  -.  (1) 


B  =  90°-A.  (2) 


(3) 


Checks :  tan  B  =  — 
a 


a 


62 


sin^         ^'  =z(c-H6)(c-6). 

»  This  is  to  be  filled  after  the  values  of  the  unknown  quantities  have  been 
found.  It  is  advisable  to  indicate  the  given  parts  and  the  unknown  parts 
clearly. 


27.]  EXAMPLES.  45 

Logarithmic  formulas  :  log  tan  ^  =  log  a  —  log  b. 
[See  Note  6] ,  log  c  =  log  a  —  log  sin  A. 

log  a  =  1.62325  log  a  =  1.62325 

log&  =  1.74819  log  sin  ^  =  9.77815 -10 

.-.  log  tan^  =  9.87506  -  10  .-.  log  c  =  1.84510 

.-.  ^  =  36°  52' 12"  .-.  c  =  70 

.-.  5  =  53°   7' 48" 

The  work  can  be  more  compactly  arranged,  as  follows : 

Checks : 
log  a  =  1 .62325  log  tan  B  =  10. 12494  -  10 

log  6  =  1. 74819  .  •.  J5  =  53°  7'  48" 


,  log  tan  A  =  9.87506  -  10  c  +  b  =  126 

.-.  ^  =  36° 52' 12"  c-b=    14 

.-.  5  =  53°   7' 48"  log  (c  + 6)  =  2.10037 

log  sin  A  =  9.77815  -  10  log  (c  -  6)  =  1.14613 


log  c  =  1.84510 
.-.  c  =  70 


•.  log  a2  =  3.24650 
.-.  loga  =  1.62325 


Note  \.  The  latter  form  is  preferable  when  all  the  parts  of  a  triangle  are 
required. 

Note  2.  If  there  is  difficulty  in  calculating  log  a  —  log  sin  A  in  the  second 
form,  wnce  log  sin  A  on  the  edge  of  a  piece  of  paper  and  place  it  immediately 
beneath  log  a. 

Note  3.    The  formula  tan  _B  =  -  can  be  used  instead  of  (2).      A  check 

then  is  A  -\:  B  =  90°.     Instead  of  (3),  one  of  the  following  formulas  can  be 
used,  viz. 

b  a  b 

sin  5'         cos  5'         cos  J. 

There  is  often  a  choice  of  formulas  that  can  be  used  in  a  solution. 

Note  4.  In  every  example  it  is  advisable  to  make  a  complete  skeleton 
scheme  of  the  solution^  before  using  the  tables  and  proceeding  with  the 
actual  computation.  In  the  last  exercise,  for  instance,  such  a  skeleton 
scheme  can  be  seen  on  erasing  all  the  numerical  quantities  in  the  equations 
that  follow  the  logarithmic  formulas. 

Note  5.  Time  will  be  saved  if  all  the  logarithms  that  can  be  found  at  one 
place  in  the  tables,  be  written  at  one  time.  Thus,  for  example,  in  the 
preceding  exercise  find  log  sin  A  immediately  after  A  has  been  found. 


46 


PLANE  TRIGONOMETRY. 


LCh.  III. 


Note  6.  The  logarithmic  formulas  can  be  written  on  a  glance  at  the 
formulas  such  as  (1),  (2),  (3).  The  writing  of  the  logarithmic  formulas 
may  be  dispensed  with  when  the  student  has  become  familiar  with  calcu- 
lation by  logarithms.  A  glance  at  the  original  formulas  will  show  how  the 
logarithms  are  to  be  combined  in  the  computation. 

2.   In  a  triangle  ABG  right  angled  at  C,  c  =  60  ft.,  6  =  50  ft. ;  find  side 
a  and  the  acute  angles. 
B 
^.^  I.    Computation  without  logarithms. 

a  cos^=-  =  — =  .8333.     .-.  ^  =  33°33^75. 

c     60 

B  =  90°  -  A.  .'.  B  -  56°  26'. 25. 


A          b~50Ft.  a                      a  =  c  sin  ^  =  60  X  .5528  =  33.17  ft. 

^^^'  I''-                 Check :  a  =  6  tan  ^  =  50  x  .6635  =  33.17. 

II.    Computation  with  logarithms. 

Given  :                c  =  60  ft. 

To  find :  A  - 

6  =  50  ft. 

B  = 

Formulas :  cos  A=  — 

a  = 

c 
B  =  90°-  A. 

Checks:  a^=c^-h'^={c-\-h){c- 
a  =  6  tan  A. 

-6). 

a  =  c  sin  A. 

, 

Logarithmic  formulas  :  log  cos  A  = 

log  h  —  log  c. 

(If  necessary.)                        log  a  = 

log  c  +  log  sin  A. 

log6  =  1.69897            (1) 

log  tan  ^  =  9.82173 -10 

(6) 

log  c  =  1.77815             (2) 

.-.  loga=  1.52070 

(7) 

.-.  log  cos  A  =  9.92082  -  10    (3) 

=  (l)  +  (6) 

=  (l)-(2) 

c-hb  =  110 

.-.  A  =  33°  33'  27" 

c-b=:    10 

.-.  B  =  56°  26'  33" 

log  (c  +  b)  =  2.04139 

log  sin  ^  =  9.74255 -10    (4) 

log(c-6)=l 

.-.log  a  =  1.52070            (5) 

.-.  log  a2  =  3.04139 

-(2) +  (4) 

.-.  loga  =  1.52070 

.-.a  =  33.10 

Note.  There  is  a  slight  difference  between  the  results  obtained  by  the 
two  methods.  This  is  due  to  the  fact  that  the  calculations  have  been  made 
with  a  four-place  table  in  one  case,  and  with  a  five-place  table  in  the  other. 
A  four-place  table  will  give  an  angle  correctly  to  within  one  minute ;  a  five- 
place  table  will  give  it  correctly  to  within  six  seconds,  and  sometimes,  to 
within  a  second. 

Ex.  Make  the  computation  I.  with  a  five-place  table. 


27.] 


EXAMPLES. 


47 


3.    In  a  triangle  right  angled  at  C,  the  hypotenuse  is  250  ft.,  and  angle  A 
is  67°  30'.     Solve  the  triangle. 

I.    Computation  without  loganthms. 
^  =  90°  -  ^  =  90°  -  67°  30'  =  22°  30'. 
a  =  c  sin  ^  =  250  X  sin  67°  30'  =  250  x  .9239  =  230.98. 
5  =  c  cos  ^  =  250  X  cos  67°  30'  =  250  x  .3827  =    95.68. 
Checks :  a^  =  (fi  —  h^,  or  a  =  b  tan  A. 


II.    Computation  with  logarithms. 
Given  :  c  =  250  ft. 

A  =  67°  30'. 


To  find  :  B  = 
a  = 
6  = 


Formulas : 


J5  =  90^  -  A. 
a  =  c  sin  A. 
b  =  G  cos  A. 


Checks :     a^ 


=  c'^-b^ 
=  (c-}-b)ic 


6). 


Logarithmic  formulas 


log  a  =  log  c  +  log  sin  A. 
log  6  =  log  c  +  log  cos  A. 


.'.  B  =  22°  30' 

log  c  =  2.39794 

log  sin  ^  =  9.96562 -10 

Jogcos^  =  9.58284-  10 

.-.log  a  =  2.36356 

.-.log  6  =  1.98078 

.-.a  =  230.97 

.-.  b  =    95.67 


c  +  6  =  345.67 
c-b  =  154.. 33 

log  (c  + 6)  =  2.53866 

log(c-  &)  =  2.18845 

.-.  log  a2  =  4.72711 

.-.log  a  =  2.36356 


\ 


4.   In  a  triangle  ABC  right  angled  at  C,  6  =  300  ft.  and  ^  =  37°  20'. 
Solve  the  triangle. 

I.    Computation  without  loganthms. 

B  =  90-A  =  90°-  37°  30'  =  52°  40'. 
b  300 


c  = 


=  377.3. 


cos^     .7951 
a  =  6  tan  u4  =  300  X  .7627  =  228.8. 
Checks :  a^  =  c^  —  b"^,  a  =  c  sin  A. 
11.    Computation  with  logarithms. 


Given 


Formulas  :    B 


37°  20'. 
300  ft. 

90°  -  A. 

cos^ 
a  =  b  tan  A. 


To  find 


Checks 


48 


PLANE  TRIGONOMETRY, 


[Ch.  III. 


.'.B  =  52°  40' 

log  6  =  2.47712 
log  cos  ^  =  9.90043 -10 
log  tan  A  =  9.88236  -  10 
.-.  log  c  =  2.57669 
.♦.  loga  =  2.35948 
.-.  c  =  377.3 
.-.  a  =  228.8 


c  +  5  =  677.3 

c-h=    77.3 

log  (c  + 6)  =  2.83078 

log  (c- 6)  =1.88818 

/.  log  a2  =  4.71896 

/.  log  a  =  2.35948 


I 


N.B.     Check  all  results  in  the  following  examples. 
belong  to  a  triangle  ABC  which  is  right  angled  at  C. 


The  given  elements 


From  the  given  elements  solve  the  following  triangles : 


6. 

c  =  18.7, 

a  =  16.98. 

6. 

a  =  194.5, 

b  =  233.5. 

7. 

c  =  2934, 

A  =  31°- 14'  12''. 

8. 

a  =  36.5, 

B  =  68°  52' 

9. 

a  =  58.5, 

b  =  100.5. 

10. 

c  =  45.96, 

a  =  1.095. 

11. 

c  =  324, 

A  =  48°  17'. 

12. 

b  =  250, 

A  =  51°  19' 

13. 

c  =  1716, 

^  =  37°  20' 30". 

14. 

a  =  2314, 

b  =  1768. 

15. 

b  =  3741, 

^  =  27°  45' 20". 

16. 

c  =  50.13, 

a  =  24.62. 

Solve  Exs.  17-24  by  two  methods,  viz.  :  (1)  with  logarithms  ; 

(2)  without  logarithms. 

^  =  62°  40'.  18.  c=9,  a  =  5. 

&  =  7.5.  20.  c  =  15,  ^  =  39°40^ 

5  =  71°  20'.  22.  c  =  12,  a  =  8. 

5  =  42°  30'.  24.  a  =  8,  6  =  12. 


17. 

a  =  40, 

19. 

a  =  4.5, 

21. 

c  =  12, 

23. 

6  =  15, 

N.B.     Questions  and  exercises  suitable  for  practice  and  review  on  the 
subject-matter  of  Chapter  III.  will  be  found  at  page  183. 


CHAPTER  IV. 

APPLICATIONS   INVOLVING  THE    SOLUTION   OF 
RIGHT-ANGLED   TRIANGLES. 

Some  practical  applications  of  trigonometry  will  now  be  given. 
It  is  not  necessary  that  all  the  problems  be  solved,  or  all  the 
articles  be  considered,  before  Chapter  V.  is  taken  up. 

28.  Projection  of  a  straight  line  upon  another  straight  line.  If 
from  a  point  P  a  perpendicular  PO  be  drawn  to  the  straight  line 
STj  then  0  is  called  the  projection  of  the  point  P  upon  the 
line  ST.     If  perpendiculars  be  drawn  from  -two  points  A,  By  to 


Fig.  20. 

a  line  LR,  and  intersect  LP  in  M,  N,  respectively,  then  MN  is 
called  the  projection  of  AB  upon  LP. 

Let  I  be  the  length  of  AB,  and  let  a  be  the  angle  at  which  the 
two  lines  AB,  LP  are  inclined  to  each  other.  Through  A  draw 
AD  parallel  to  LP.     Then 

Projection  =  MN=  AD  =  AB  cos  DAB  =  I  cos  a. 

That  is,  the  projection  of  one  straight  line  upon  another  straight 
line  is  equal  to  the  product  of  the  length  of  the  first  line  and  the 
cosine  of  the  angle  of  inclination  of  the  two  lines. 

Note.  The  projection  discussed  here,  is  orthogonal  (i.e.  perpendicular) 
projection.  If  a  pair  of  parallel  lines  AMi^  BNi^  not  perpendicular  to  Zi?, 
be  drawn  through  J.,  B^  then  M^Ni  is  an  oblique  projection  of  AB  on  LB. 

49  _ 


50  PLANE  TBIGONOMETRT.  [Ch.  IV. 

EXAMPLES. 

In  working  these  examples  use  logarithms  or  not,  as  appears  most  con- 
venient.    Check  the  results. 

1.  A  ladder  28  ft.  long  is  leaning  against  the  side  of  a  house,  and  makes 
an  angle  27°  with  the  wall.  Find  its  projections  upon  the  wall  and  upon  the 
ground. 

2.  What  is  the  projection  of  a  line  87  in.  long  upon  a  line  inclined  to  it  at 
an  angle  47°  30'  ? 

3.  What  are  the  projections  :  (a)  of  a  line  10  in.  long  upon  a  line  inclined 
22°  30'  to  it  ?  (6)  of  a  line  27  ft.  6  in.  long  upon  a  line  inclined  37°  to  it  ? 
(c)  of  a  line  43  ft.  7  in.  long  upon  a  line  inclined  67°  20'  to  it  ?  (d)  of  a  line 
34  ft.  4  in.  long  upon  a  line  inclined  55°  47'  to  it  ? 

29.  Measurement  of  heights  and  distances.  There  are  various 
instruments  used  for  measuring  angles.  The  sextant  can  be  used 
for  measuring  the  angle  between  the  two  lines  drawn  from  the 
observer's  eye  to  each  of  two  distant  objects.  Horizontal  and 
vertical  angles  can  be  measured  with  a  theodolite  or  engineer's 
transit.  When  great  accuracy  is  not  required,  vertical  angles  can 
be  measured  by  means  of  a  quadrant. 

When  an  object  is  above  the  observer's  eye,  the  angle  between 
the  line  from  the  eye  to  the  object,  and  the  horizontal  line 
through  the  eye  and  in  the  same  vertical  plane  as  the  first  line, 
is  called  the  angle  of  elevation  of  the  object,  or  simply  the  elevation 
of  the  object.  When  the  object  is  below  the  observer's  eye,  this 
angle  is  called  the  angle  of  depression  of  the  object,  or  simply  Hie 
depression  of  the  object. 


Fig.  21. 

Note.  In  ordinary  work  engineers  get  angular  measurements  exact  to 
within  one  minute,  and  in  the  best  ordinary  work  to  half  a  minute.  In 
very  particular  work,  like  geodetic  survey,  they  can  get  measurements  exact 
to  five  seconds.  For  ordinary  engineering  work  five-place  tables  are  gen- 
erally used  ;  four-place  tables  are  used  in  some  kinds  of  work.  See  Art.  11, 
Note  1,  Art.  27,  Ex.  2,  Note. 


>.] 


HEIGHTS  AND  DISTANCES. 


51 


EXAMPLES. 


A  few  simple  examples  are  given  here  ;  others  ivill  be  given  later. 

1.  At  a  point  150  ft.  from,  and  on  a  level  with,  the 
base  of  a  tower,  the  angle  of  elevation  of  the  top  of 
the  tower  is  observed  to  be  60°.  Find  the  height  of 
the  tower. 

Let  AB  be  the  tower,  and  P  the  point  of  observa- 
tion. 

By  the  observations, 

^lP=150ft.,   APB  =  e()°. 

^P=^Ptan60°=150xV3  =  150x  1.7321=279.82  ft. 

2.  In  order  to  find  the  height  of  a  hill,  a  line  was  measured  equal  to 
100  ft.,  in  the  same  level  with  the  base  of  the  hill,  and  in  the  same  vertical 
plane  with  its  top.  At  the  ends  of  this  line  the  angles  of  elevation  of  the  top 
of  the  hill  were  30°  and  45°.     Find  the  height  of  the  hill. 

Let  P  be  the  top  of  the  hill,  and  AB  the  base  line.    The  vertical  line 

through  P  will  meet  AB  produced  in  C. 
AB  =  100  ft.,  C^P=  30°,  OPP=45°  j  the 
height  CP  is  required.  Let  BG  =  x^  and 
GP=y. 

In  triangle  CAP, 

CP, 
^     AG' 
CP 


tan  30°; 


l-^-lOO  Ft 


Fig.  23. 


in  GBP, 


BC 


=  tan  45°. 


Hence, 
and 


y. —  =  tan  30°  =  .57735, 

a;  +  100 

^  =  tan  45°  =  1. 

X 


(1) 
(2) 


From  (2),  x  =  y.     Substitution  in  (1)  gives 

2/=(?/  +  100)x.67735. 
.-.  2/(1 -.57735)  =57.735. 

.-.  CP^  2/  =  ^^:^=  136.6  ft. 
^      .42265 

3.  A  flagstaff  30  ft.  high  stands  on  the  top  of  a  cliff,  and  from  a  point  on 
a  level  with  the  base  of  the  cliff  the  angles  of  elevation  of  the  top  and  bottom 
of  the  flagstaff  are  observed  to  be  40°  20'  and  38°  20',  respectively.  Find  the 
height  of  the  cliff. 


52 


PLANE  TRIGONOMETRY, 


[Cn.  IV. 


Let  BP  be  the  flagstaff  on  the  top  of  the  cliff  BL,  and  let  C  be  the  place 
of  observation.     i?P  =  30  ft.,  LGB  =  38°  20',  LCP  =  40°  20'.     Let  CL  =  x, 
LB  =  y. 


In  LCB, 


In  LCP, 


LB 


=  tan  38°  20' 


tan  40°  20' 


Hence,  on  division, 
On  solving  for  y, 


LG 

y-  =  .7907. 

X 

LP 
LG 

y±M  =  .8491. 

X 

y      ^7907 
y  +  30      8491* 

LB  =  y  =  4SiQ.\?>  it. 


4."  At  a  point  180  ft.  from  a  tower,  and  on  a  level  with  its  base,  the  eleva- 
tion of  the  top  of  the  tower  is  found  to  be  65°  40. 5'.  What  is  the  height  of 
the  tower  ? 

5.  From  the  top  of  a  tower  120  ft.  high  the  angle  of  depression  of  an 
object  on  a  level  with  the  base  of  the  tower  is  27°  43'.  What  is  the  distance 
of  the  object  from  the  top  and  bottom  of  the  tower  ? 

6.  From  the  foot  of  a  post  the  elevation  of  the  top  of  a  column  is  45°,  and 
from  the  top  of  the  post,  which  is  27  ft.  high,  the  elevation  is  30°.  Find  the 
height  and  distance  of  the  column. 

7.  From  the  top  of  a  cliff  120  ft.  high  the  angles  of  depression  of  two 
boats,  which  are  due  south  of  the  observer,  are  20°  20'  and  68°  40'.  Find  the 
distance  between  the  boats. 

8.  From  the  top  of  a  hill  450  ft.  high,  the  angle  of  depression  of  the  top 
of  a  tower,  which  is  known  to  be  200  ft.  high,  is  63°  20'.  What  is  the  distance 
from  the  foot  of  the  tower  to  the  top  of  the  hill  ? 

9.  From  the  top  of  a  hill  the  angles  of  depression  of  two  consecutive 
mile-stones,  which  are  in  a  direction  due  east,  are  21°  30'  and  47°  40'.  How 
high  is  the  hill  ? 

10.  For  an  observer  standing  on  the  bank  of  a  river,  the  angular  elevation 
of  the  top  of  a  tree  on  the  opposite  bank  is  60° ;  when  he  retires  100  ft.  from 
the  edge  of  the  river  the  angle  of  elevation  is  30°.  Find  the  height  of  the  tree 
and  the  breadth  of  the  river. 

11.  Find  the  distance  in  space  travelled  in  an  hour,  in  consequence  of  the 
earth's  rotation,  by  an  object  in  latitude  44°  20'.  [Take  earth's  diameter 
equal  to  8000  mi.] 

12.  At  a  point  straight  in  front  of  one  corner  of  a  house,  its  height  subtends 
an  angle  34°  45',  and  its  length  subtends  an  angle  72°  30'  j  the  height  of  the 
house  is  48  ft.     Find  its  length. 


30.] 


THE  MARINER  S   COMPASS. 


53 


30.  Problems  requiring  a  knowledge  of  the  points  of  the  Mariner's 
Compass.  The  circle  in  the  Mariner's  Compass  is  divided  into  32 
equal  parts,  each  part  being  thus  equal  to  360°  -r-  32,  i.e. 
The  points  of  division  are  named  as  indicated  on  the  tigure. 


Hi' 


Fig.  25. 

It  will  be  observed  that  the  points  are  named  with  reference  to 
the  points  North,  South,  East,  and  West,  which  are  called  the 
cardinal  points.  Direction  is  indicated  in  a  variety  of  ways. 
For  instance,  suppose  C  were  the  centre  of  the  circle ;  then  the 
point  P  in  the  figure  is  said  to  bear  E.N.E.  from  C,  or,  from  C  the 
hearing  of  P  is  E.N.E,  Similarly,  C  bears  W.S.W.  from  P,  or, 
the  bearing  of  C  from  P  is  W.S.W.  The  point  E.N.E.  is  2  poii;ts 
North  of  East,  and  6  points  East  of  North.  Accordingly,  the 
phrases  E.  22i°  N.,  N.  67i°  E.,  are  sometimes  used  instead  of 
E.N.E. 

EXAMPLES. 

1.  Two  ships  leave  the  same  dock  at  8  a.m. 
in  directions  S.W.  by  S.,  and  S.E.  by  E.  at 
rates  of  9  and  9|  mi.  an  hour  respectively. 
Find  their  distance  apart,  and  the  bearing  of 
one  from  the  other  at  10  a.m.  and  at  noon. 

2.  From  a  lighthouse  L  two  ships  A  and  B 
are  observed  in  a  direction  N.E.  and  N.  20°  "W. 
respectively.  At  the  same  time  A  beai's  S.E. 
from  B.     If  LA  is  6  mi.,  what  is  LB  ? 


54 


PLANE  TRIGONOMETRY. 


[Ch.  IV. 


31.  Mensuration.  Let  ABC  be  any  triangle,  and  let  the  lengths 
of  the  sides  opposite  the  angle  A,  B,  C  be  denoted  by  a,  6,  c,  respec- 
tively.   From  any  vertex  C  draw  CD  at  right  angles  to  the  opposite 


side  AB.  It  has  been  shown  in  arithmetic  and  geometry,  that  the 
area  of  a  triangle  is  equal  to  one-half  the  product  of  the  lengths 
of  any  side  and  the  perpendicular  drawn  to  it  from  the  opposite 
vertex.     [In  (1)  A  is  acute,  in  (2)  A  is  obtuse.] 

area  ABC  (Tig.  l)=iAB'  DC; 

=  iAB- AC  sin  A; 
=  ibc  sin  A. 

area  ABC  (Fig.  2)  =  iAB'DC; 

=  iAB'ACsmCAD; 
=  ibc  sin  (180  -  A). 

It  will  be  seen  in  Art.  45,  that  sin  (180  —  A)  =  sin  A.  Hence, 
the  area  of  a  triangle  is  equal  to  one-half  the  product  of  any  two  sides 
and  the  sine  of  their  contained  angle. 


EXAMPLES. 

1.  Find  the  area  of  the  triangle  in  which  two  sides  are  31  ft.  and  23  ft. 
and  their  contained  angle  67°  30'. 

area  =  §l-><^  X  sin  67°  30'  =  31  x  23  x  .92388  ^  339  37        f ^^ 
2  2 

2.  Find  area  of  triangle  having  sides  125  ft.,  80  ft.,  contained  angle  28°  36'. 

3.  Find  area  of  triangle  having  sides  125  ft.,  80  ft.,  contained  angle  151°  25'. 
[Draw  figures  carefully  for  Exs.  2,  3.] 

4.  Find  area  of  parallelogram  two  of  whose  adjacent  sides  are  243,  315  yd., 
and  their  included  angle  35°  40'. 


32.J  ISOSCELES   TiiiANGLES.  55 

5.  Find  area  of  parallelogram  two  of  whose  adjacent  sides  are  14,  15  ft., 
and  included  angle  Ib"^. 

6.  Find  area  of  triangle  having  sides  40  ft.,  45  ft.,  with  an  included  angle 
28°  57'  18". 

7.  Write  two  other  formulas  for  area  ABC,  similar  to  that  derived  above. 
Also,  derive  them. 

32.  Solution  of  isosceles  triangles.  In  an  isosceles  triangle,  the 
perpendicular  let  fall  from  the  vertex  to  the  base  bisects  the  base 
and  bisects  the  vertical  angle.  An  isosceles  triangle  can  often  be 
solved  on  dividing  it  into  two  equal  right-angled  triangles. 

EXAMPLES. 

1.  The  base  of  an  isosceles  triangle  is  24  in.  long,  and  the  vertical  angle  is 
48°  ;  find  the  other  angles  and  sides,  the  perpendicular  from  the  vertex  and  the 
area.    Only  the  steps  in  the  solution  will  be  indicated. 

Let  ABC  be  an  isosceles  triangle  having  base  AB  =  24 

in.,  angle  C  =  48°.     Draw  CD  at  right  angles  to  base  ; 

then  CD  bisects  the  angle  ACB  and  base  AB.     Hence,  y 

in  the  right-angled  triangle  ADC,  AD  —  \  AB  =  12,  / 

ACD  =  I  ACB  =  24°.      Hence,   angle  A,   sides  AC,  / 

DC,  and  the  area,  can  be  found.  / 

2.  In  an  isosceles  triangle  each  of  the  equal  sides  is         s~-i2-"- 
363  ft.,  and  each  of  the  equal  angles  is  75°.     Find  the  p  «  28 
base,  perpendicular  on  base,  and  the  area. 

3.  In  an  isosceles  triangle  each  of  the  equal  sides  is  241  ft.,  and  their 
included  angle  is  96°.    Find  the  base,  angles  at  the  base,  height,  and  area. 

4.  In  an  isosceles  triangle  the  base  is  65  ft. ,  and  each  of  the  other  sides  is 
90  ft.     Find  the  angles,  height,  and  area. 

6.  In  an  isosceles  triangle  the  base  is  40  ft.,  height  is  30  ft.  Find  sides, 
angles,  area. 

6.  In  an  isosceles  triangle  the  height  is  60  ft.,  one  of  equal  sides  is  80  ft. 
Find  base,  angles,  area. 

7.  In  an  isosceles  triangle  the  height  is  40  ft.,  each  of  equal  angles  is  63°. 
Find  sides  and  area. 

8.  In  an  isosceles  triangle  the  height  is  63  ft.,  vertical  angle  is  75°.  Find 
sides  and  area. 

33.  Related  regular  polygons  and  circles.  The  knowledge  of 
trigonometry  thiis  far  attained,  is  of  service   in  solving  many 


56  PLAKB  TJUQONOMETHY.  [Ch.  IV. 

problems  in  which  circles  and  regular  polygons  are  concerned. 
Some  of  these  problems  are : 

(a)  Given  the  length  of  the  side  of  a  regular  polygon  of  a  given 
number  of  sides,  to  find  its  area;  also,  to  find  the  radii  of  the 
inscribed  and  circumscribing  circles  of  the  polygon ; 

(b)  To  find  the  lengths  of  the  sides  of  regular  polygons  of  a 
given  number  of  sides  which  are  inscribed  in,  and  circumscribed 
about,  a  circle  of  given  radius. 


J^IQ.  29. 


Eor  example,  let  AB  (Fig.  29)  be  a  side,  equal  to  2  a,  of  a  regu- 
lar polygon  of  n  sides,  and  let  C  be  the  centre  of  the  inscribed 
circle.  Draw  CA,  CB,  and  draw  CD  at  right  angles  to  AB.  Then 
D  is  the  middle  point  of  AB. 

360°     180° 


By  geometry,  angle  ACD  =  ^  ACB  =  i 

It  IV 

Also,  by  geometry, 

Tangle  DAO  =  ^(^^^^0^  =  f^)^*^"-! 

Hence,  in  the  triangle  ADC,  the  side  AD  and  the  angles  are 
known ;  therefore  CD,  the  radius  of  the  circle  inscribed  in  the 
polygon,  can  be  found.  On  making  similar  constructions,  the 
solution  of  the  other  problems  referred  to  above  will  be  apparent. 
The  perpendicular  from  the  centre  of  the  circle  to  a  side  of  the 
inscribed  polygon  is  called  the  apothem  of  the  polygon. 

EXAMPLES. 

1.  The  side  of  a  regular  heptagon  is  14  ft.  :  find  the  radii  of  the  inscribed 
and  circumscribing  circk\s  ;  also,  find  the  difference  between  the  areas  of  the 
heptagon  and  the  inscribed  circle,  and  the  difference  between  the  area  of 
the  heptagon  and  the  area  of  the  circumscribing  circle. 


34.]  OBLIQUE  TRIANGLES.  67 

2.  The  side  of  a  regular  pentagon  is  24  ft.    Find  quantities  as  in  Ex.  1. 

3.  The  side  of  a  regular  octagon  is  24  ft.     Find  quantities  as  in  Ex.  1. 

4.  The  radius  of  a  circle  is  24  ft.  Find  the  lengths  of  the  sides  and  apo- 
thems  of  the  inscribed  regular  triangle,  quadrilateral,  pentagon,  hexagon, 
heptagon,  and  octagon.  Compare  the  area  of  the  circle  and  the  areas  of 
these  regidar  polygons  ;  also  compare  the  perimeters  of  the  polygons  and 
the  circumference  of  the  circle. 

5.  For  the  same  circle  as  in  Ex.  4,  find  the  lengths  of  the  sides  of  the 
circumscribing  regular  figures  named  in  Ex.  4.  Compare  their  areas  and 
perimeters  with  the  area  and  circumference  of  the  circle. 

6.  If  a  be  the  side  of  a  regular  polygon  of  n  sides,  show  that  i?,  the 

180° 
radius  of  the  circumscribing  circle,  is  equal  to  ^  a  cosec ;  and  that  r,  the 

180°       ^ 

radius  of  the  circle  inscribed,  is  equal  to  A  a  cot 

n 

7.  If  r  be  the  radius  of  a  circle,  show  that  the  side  of  the  regular  inscribed 

180° 
polygon  of  n  sides  is  2  r  sin ;  and  that  the  side  of  the  regular  circum- 

180°    ^ 

scribing  polygon  is  2  r  tan . 

n 

8.  If  a  be  the  side  of  a  regular  polygon  of  n  sides,  B  the  radius  of  the  cir- 
cumscribing circle,  and  r  the  radius  of  the  circle  inscribed,  show  that  area  of 

polygon  =  i  naP'  cot  —  =  1  nija  sin  '^  =  nr^  tan  — • 
n  n  n 

34.  Solution  of  oblique  triangles.  Since  an  oblique  triangle  can 
be  divided  into  right-angled  triangles  by  drawing  a  perpendicular 
from  a  vertex  to  the  opposite  side,  it  may  be  expected  that  know- 
ledge concerning  the  solution  of  right-angled  triangles  will  be  of 
service  in  solving  oblique  triangles.  This  expectation  will  not  be 
disappointed.  An  examination,  which  it  is  advisable  for  the  stu- 
dent to  make  before  proceeding  farther,  will  show  that  all  the  sets 
of  data  from  which  a  definite  triangle  can  be  drawn  are  those 
indicated  in  (l)-(4)  below.  The  ability  to  make  the  following  geo- 
metrical constructio7is  is  presupposed : 

(1)  To  draw  a  triangle  on  being  given  two  of  its  angles  and  a  side  oppo- 
site to  one  of  them  ; 

(2)  To  draw  a  triangle  on  being  given  two  of  its  sides  and  an  angle  oppo- 
site to  one  of  them  ; 

(3)  To  draw  a  triangle  on  being  given  two  of  its  sides  and  their  included 
angle ; 

(4)  To  draw  a  triangle  on  being  given  its  three  sides. 


58 


PLANE  TRIGONOMETRY, 


[Ch.  IV. 


In  what  follows,  only  the  steps  in  the  solutions  will  be  indi- 
cated. The  examples  that  are  worked  may  be  saved,  so  that  the 
amount  of  labor  required  by  the  method  of  solution  shown  here 
can  be  compared  with  the  amount  required  by  another  method 
which  will  be  described  later. 

There  are  four  cases  in  the  solution  of  oblique  triangles ;  these 
cases  correspond  to  the  four  problems  of  construction  stated  above. 

Case  I.    Given  tioo  angles  and  a  side  opposite  to  one  of  them. 

o  c 


Fig.  30. 

In  ABG  let  A,  B,  a  be  known.     Angle  C  and  sides  6,  c  are  re« 
quired.     From  G  draw  CD  at  right  angles  to  AB  or  AB  produced. 

In  triangle  CBD,  angle  CBD  and  side  CB  are  known.     .-.  BD 
and  DC  can  be  found. 

Then,   in  triangle  ACD^  side  DG  and  angle  A  are  known. 
.'.AG  and  AD  can  be  found. 

Side  AB  =  AjD-BD  when  B  is  obtuse,  smd  AB  =  AD  +  DB 
when  B  is  acute. 

Angle  G=1S0°-(A  +  B). 

Another  method  of  solution  is  given  in  Art.  55. 


1.  Ex.  1,  Art.  55. 
3.   Ex.  2,  Art.  60. 


EXAMPLES. 

2.  Ex.  2,  Art.  55. 

4.   Other  Exs.  in  Arts.  65,  60. 


Case  II.    Given  two  sides  and  an  angle  opposite  to  one  of  them. 

N.B.    The  first  part  of  the  text  in  Art.  56  should  be  read  at  this  time. 

Let  (Fig.  30)  AG,  BG,  angle  A  be  known.  [In  a  certain  case, 
as  shown  in  Art.  56,  two  triangles  can  be  drawn  which  satisfy 
the  given  conditions.]  From  G  draw  CD  at  right  angles  to  AB 
or  AB  produced. 


OnLiqUM  TMtANGLES,  59 

In  ACB,  AC  and  A  are  known.     .-.  AD,  DC,  angle  ACD,  can 

be  found- 
Then,  in  BCD,  BC  and  CD  are  known.     .-.  BD,  angle  DBC, 

can  be  found. 

In  one  figure,  AB  =  AD-  BD,  angle  ABC  =  180°  -  CBD. 
In  other  figure,  AB  =  J.Z>  +  DB.     In  both  figures, 

angle  ACB  =  180°  -  (CAB  +  ^5(7). 

Another'  method  of  solution  is  given  in  Art.  ^Q, 

EXAMPLES. 
1.  Ex.  1,  Art.  56.  2.  Ex.  2,  Art.  56. 

3.  Ex.  1,  Art.  60.  4,   Other  Exs.  in  Arts.  66,  60. 

Case  III.    Given  two  sides  and  their  included  angle. 

In  ABC  let  b,  c,  A  be  known.  Side  a,  B,  C  are  required. 
From  C  draw  CD  at  right  angles  to  AB  or  AB  produced. 

In  ACD,  AC  and  angle  A  are  known.  .-.  CD  and  AD  can  be 
found. 

Then,  in  triangle  CDB,  CD  is  now  known,  and  BD=AD—AB 
or  ^5  -  AD.  .-.  Angle  C5i>  can  be  found.  Angle  ABC  =  180° 
-  CBD  in  figure  on  the  left.     Angle  ACB  =  180°  -{A  +  B). 

Another  method  of  solution  is  given  in  Art.  57. 

EXAMPLES. 

1.  Ex.  1,  Art.  57.  2.   Ex.  2,  Art.  67.  3.   Ex.  1,  Art.  61. 

4.  Other  Exs.  in  Arts.  57,  61.  6,   Ex.  in  Art.  21. 

Case  IV.    Given  the  three  sides. 

In  ABC  let  a,  6,  c  be  known.  The  angles  A,  B,  C  are  required. 
From  any  vertex  C  draw  CD  at  right  angles  to  AB  or  AB  pro- 
duced. 

CD'=b'-AD'',  (1) 

also,  CD'  =  a^  -  DjB^  =  a^  -  (c  -  ADf. 

.-.  62  _  ^/>2  ^a'-(c-  ADf. 

.%^Z>  =  ^'  +  ^'-<  (2) 

2c  ^  '^ 


60  PLANE  TRIGONOMETRY,  [Ch.  IV. 

Also,  DB  =  AD  —  c  (one  figure),  or  c  —  AD  (other  figure). 
Hence,  in  ACD,  AC,  AD  are  known.     .-.  A  can  be  found. 
Also,  in  CDB,  CB,  DB  are  known.     .-.  B  can  be  found. 

C  =  1S0°-(A  +  B), 

Another  method  of  solution  is  given  in  Art.  58. 

EXAMPLES. 

1.   Ex.  1,  Art.  58.  2,  Ex.  2,  Art.  58. 

3.  Ex.  1,  Art.  62.  4.  Other  Exs.  in  Arts.  58,  62. 

5.  Solve  some  of  the  problems  in  Art.  63  by  means  of  right-angled 
triangles. 

34  a.  The  area  of  a  triangle  in  terms  of  the  sides.    (See  Fig.  30.) 
From  (1),  (2),  Case  IV.,  Art.  34, 

CT2  _  ^2      fh'  +  c^  -  ay^  4  c'h'  -  Qf-  +  c^  -  a^ 
\       2c       J  4c2 

_  [2  c6  +  ?>'  +  c^  -  a^M2  ch  -  (b'  +  c^  -  a')1 
4c2 

_  [(6  +  cY  -  g^  [«'  -  (^  -  c)n 
4c2 

^  (g  +  6  +  c)(-  g  +  5  +  c)(a  -  &  +  c)(a  +  6  -  c) 
4c2 

Let  a  +  b  +  c  =  2s; 

then  2(s  —  g)  =  g4-&  +  c  —  2g  =  —  g  +  6+c. 

Similarly,  2  (s  —  b)  =  a —  b -\-  c;  2  (s  —  c)  =  a  +  b  —  c. 
2 


I 


Then  CD  =  -  V<s  -  g)(s  -  6)(s  -  c). 

c 


.-.  Area  ^J5(7  =\AB'CD  =  Vs(s  -  g)(s  -  6)(s  -  c)* 

Ex.  Eind  the  areas  of  the  triangles  in  Exs.  Case  IV.,  Art.  34.  Check  the 
results  by  finding  the  areas  by  the  method  of  Art.  31. 

*  This  is  sometimes  known  as  Ilero'^s  Formula  for  the  area  of  a  triangle. 
It  was  discovered  by  Hero  (or  Heron)  of  Alexandria,  who  lived  about 
125  B.C.,  and  placed  engineering  and  land  surveying  on  a  scientific  basis. 


p       II 


U  b;  34  c]  DISTANCE  OF  VISIBLE  BOUIZON.  61 

34  b.   Distance  and  dip  of  the  visible  horizon. 

Let  C  be  the  centre  of  the  earth,  and  let  the  radius  be  denoted  by  r. 

Let  P  be  a  point  above  the  eartli's  surface,  and  let 
its  height  PL  be  denoted  by  h. 

Join  P,  C ;  draw  PB  from  P  to  any  point  in  the 
visible  horizon  ;  draw  the  horizontal  line  PH  in  the 
same  plane  with  PC,  PB.  Then  angle  HPB  is  called 
the  dip  of  the  horizon.    By  geometry, 

angle  PPO  =  90°. 

PP2  =  P02  -  CB^  =  (ir  +  hy-r'^  =  2rh  +  h\ 

Since  h^  is  very  small  compared  with  2  7% 

PB  =  V'2  rh  approximately. 

Take  r  =  3960  mi. ,  and  let  h  be  measured  in  feet.     Then  =  height 


Fig.  30a. 


of  P  in  miles. ^^^^ 

.-.  PB  =a/2  X  3960  x-^mi.  =  V¥h  mi. 
^  6280  ^ 

Hence,  the  distance  of  the  horizon  in  miles  is  approximately  equal  to 
the  square  root  of  one  and  one-half  times  the  height  in  feet. 

EXAMPLES. 

1.  A  man  whose  eye  is  6  ft.  from  the  ground  is  standing  on  the  sea- 
shore.    How  far  distant  is  his  horizon  ? 

Distance  =  Vf  x  6  mi.  =  3  mi. 

2.  Find  the  greatest  distance  at  which  the  lamp  of  a  lighthouse  can  be 
seen,  the  light  being  80  ft.  above  the  sea  level. 

3.  Find  the  height  of  the  lamp  of  a  lighthouse  above  the  sea  level  when 
it  begins  to  be  seen  at  a  distance  of  12  mi. 

4.  From  the  top  of  a  cliff,  40  ft.  above  the  sea  level,  the  top  of  a  steamer's 
funnel  which  is  known  to  be  30  ft.  above  the  water  is  just  visible.  What 
is  the  distance  of  the  steamer  ? 

5.  Find  the  distance  and  dip  of  the  horizon  at  the  top  of  a  mountain 
3000  ft.  high? 

6.  Find  the  distance  and  dip  of  the  horizon  at  the  top  of  a  mountain 
2 1  mi.  high. 

34  c.  Examples  in  the  measurement  of  land.  In  order  to  find 
the  area  of  a  piece  of  ground,  a  surveyor  measures  distances  and 
angles  sufficient  to  provide  data  for  the  computation.    An  account 


&1 


Plane  TitiGONoMETkY. 


[Ch.  iv.j 


of  his  method  of  doing  this,  and  of  his  arrangement  of  the  data  ^^i 
and  the  results  in  a  simple,  clear,  and  convenient  form,  belongs 
to  special  works  on  surveying.  This  article  merely  gives  some 
examples  which  can  be  solved  without  any  knowledge  of  profes- 
sional details.  The  various  rules  for  finding  the  area,  of  a  triangle 
and  a  trapezoid,  are  supposed  to  be  known.  In  solving  these 
problems,  the  student  should  make  the  plotting  or  mapping  an 
important  feature  of  his  work. 

The  Gunter's  chain  is  generally  used  in  measuring  land.     It  is 
4  rods  or  66  feet  in  length,  and  is  divided  into  100  links. 

An  acre  =  10  square  chains =4  roods  =  160  square  rods  or  poles. 
The  points  of  the  compass  have  been  explained  in  Art.  30. 


EXAMPLES. 

1.   A  surveyor  starting  from  a  point  A  runs  S.  70°  E.  20  chains,  thence 
N.  10°  W.  20  chains,  thence  N.  70°  W.  10  chains,  thence  S.  20°  W.  17.32  chains 
to  the  place  of  beginning.     What  is  the  area  of 
the  field  which  he  has  gone  around  ? 

Make  a  plot  or  map  of  the  field,  namely, 
ABCD.  Here,  AB  represents  20  chains,  and 
the  bearing  of  B  from  A  is  S.  70°  E.  5(7  repre- 
sents 20  chains,  and  the  bearing  G  from  B  is 
N.  10°  W.,  and  so  on.  Through  the  most  west- 
erly point  of  the  field  draw  a  north-and-south 
line.  This  line  is  called  the  meridian.  In  the 
case  of  each  line  measured,  find  the  distance  that 
one  end  of  the  line  is  east  or  west  from  the  other 
end.  This  easting  or  westing  is  called  the  depar- 
ture of  the  line.  Also  find  the  distance  that  one 
end  of  the  line  is  north  or  south  of  the  other  end. 
This  northing  or  southing  is  called  the  latitude 
of  the  line.  For  example,  in  Fig.  30  &,  the  de- 
partures of  AB,  BC,  GD,  DA,  are  BiB,  BL,  CH,  DDu  respectively ;  the 
latitudes  of  the  boundary  lines  are  ABi,  BiGi,  C\Di,  D]_A,  respectively. 
It  should  be  observed  (Art,  36)  that  the  algebraic  sum  of  the  departures  of 
the  boundary  lines  is  zero,  and  so  also  is  the  algebraic  sum  of  their  latitudes. 
The  following  formulas  are  easily  deduced  : 

Departure  of  a  line  =  length  of  line  x  sine  of  the  bearing  ; 
Latitude  of  a  line  =:  length  of  line  x  cosine  of  the  bearing. 

By  means  of  the  departures,  the  meridian  distance  of  a  point  (i.e.  its 
distance  from   the  north-and-south  line)  can  be  found.     Thus  the  meridian 


Fig.  sod. 


35.]  SUMMARY.  63 

distance  of  C  is  CiC,  and  CiC  =  DiD  +  HC.     Hence  in  Fig.  30  &,  ^^i,  BiB, 
BiCu  CiC,  OiZ>i,  DiD  can  be  computed.    Now 

area  ABCD  =  trapezoid  D^DCCi  +  trapezoid  CiCBBi  —  triangle  ADDi 

—  triangle  ABBi. 

The  areas  in  the  second  member  can  be  computed  ;  it  will  be  found  that 
area  ABCD  =  26  acres. 

Note.  Sometimes  the  bearing  and  length  of  one  of  the  lines  enclosing  the 
area  is  also  required.  These  can  be  computed  by  means  of  the  latitudes  and 
departures  of  the  given  lines.  The  formulation  of  a  simple  rule  for  doing 
this  is  left  as  an  exercise  to  the  student. 

2.  In  Ex.  1,  deduce  the  length  and  bearing  of  DA  from  the  lengths  and 
bearings  of  AB,  BC,  CD, 

3.  A  surveyor  starts  from  A  and  runs  4  chains  S.  45°  E.  to  5,  thence  5 
chains  E.  to  (7,  thence  6  chains  N.  40°  E.  to  D.  Find  the  distance  and  bear- 
ing of  A  from  D ;  also,  the  area  of  the  field  ABCD.  Verify  the  results 
by  going  around  the  field  in  the  reverse  direction,  and  calculating  the  length 
and  bearing  of  BA  from  the  lengths  and  directions  of  AD,  DC,  CB. 

4.  A  surveyor  starts  from  one  corner  of  a  pentagonal  field,  and  runs 
N.  25°  E.  433  ft.,  thence  N.  76°  55'  E.  191  ft.,  thence  S.  6°  41'  W.  539  ft., 
thence  S.  25°  W.  40  ft.,  thence  N.  65°  W.  320  ft.  Find  the  area  of  the  field. 
Deduce  the  length  and  direction  of  one  of  the  sides  from  the  lengths  and 
directions  of  the  other  four. 

5.  From  a  station  within  a  hexagonal  field  the  distances  of  each  of  its 
corners  were  measured,  and  also  their  bearings ;  required  its  plan  and  area, 
the  distances  in  chains  and  the  bearings  of  the  corners  being  as  follows  :  7.08 
N.E.,  9.57  N.  iE.,  7.83  N.W.  by  W.,  8.25  S.W.  by  S.,  4.06  S.S.E.  7°E., 
5.89  E.  by  S,  3i°E. 

35.  Summary.  Chapter  II.  was  concerned  with  defining  and 
investigating  certain  ratios  inseparably  connected  with  (acute) 
angles,  and  attention  was  directed  to  the  tables  of  these  ratios 
and  their  logarithms.  In  Chap.  III.  it  was  shown  how  these 
definitions  and  tables  can  be  used  in  finding  parts  of  a  right- 
angled  triangle  when  certain  parts  are  known.  In  Chap.  IV. 
the  knowledge  gained  in  Chap.  III.  was  employed  in  the  solu- 
tion of  some  of  the  many  problems  in  which  right-angled  triangles 
appear.  In  Art.  34  it  has  been  seen  that  this  knowledge  can 
serve  for  the  solution  of  oblique  triangles.  It  follows,  then,  that 
it  can  serve  for  the  solution  of  problems  in  which  oblique  tri- 
angles appear,  and,  accordingly,  for  the  solution  of  all  problems 


64  PLANE   TRIGONOMETRY.  [Cn.  IV. 

invoh)iny  the  measurement  of  straight  lines  only.  Consequently, 
the  student  is  now  able,  without  any  additional  knowledge  of 
trigonometry,  to  solve  the  numerical  problems  in  Chaps.  VII., 
VIII.  It  is  thus  apparent  that  even  a  slight  acquaintance  with 
the  ratios  defined  in  Chap.  II.  has  greatly  increased  the  learner's 
ability  to  solve  useful  practical  problems. 

Oblique  triangles  can  sometimes  be  solved  in  a  more  elegant 
manner  than  that  pointed  out  in  Art.  34.  In  order  to  show  this, 
further  consideration  of  angles  and  the  trigonometric  ratios  is 
necessary.  Consequently,  in  Chap.  V.  some  important  additions 
are  made  to  the  idea  of  a  straight  line  and  the  idea  of  an  angle ; 
the  trigonometric  ratios  are  defined  in  a  more  general  way,  namely, 
for  all  angles,  instead  of  for  acute  angles  only,  and  the  principal 
relations  of  these  ratios  are  deduced.  Chapter  VI.  treats  of  the 
ratios  of  two  angles  in  combination.  While  it  is  necessary  to  con- 
sider these  matters  before  proceeding  to  the  solution  of  oblique 
triangles  given  in  Chap.  VII.,  it  should  be  said  that  the  know- 
ledge that  will  be  gained  in  Chaps.  V.,  VI.,  VII.,  is  necessary  and 
important  for  other  purposes  besides  the  solution  of  triangles. 
In  fact,  the  latter  is  one  of  the  least  important  of  the  results 
obtained  in  these  chapters. 

N.B.  Questions  and  exercises  suitable  for  practice  and  review  on  the  sub- 
ject-matter of  Chap.  IV.  will  be  found  at  page  184. 


CHAPTER   V. 

TRIGONOMETRIC    RATIOS    OF    ANGLES    IN    GENERAL. 

36.  Directed  lines.  Let  MN  be  a  line  unlimited  in  length  in 
the  directions  of  both  M  and  N.  Suppose  that  a  point  starts  at 
P  and  moves  along  this  line  for  some  given  distance.  In  order 
to  mark  where  the  point  stops,  it  is  necessary  to  know,  not  only 
this  distance,  but  also  the  direction  in  which  the  point  has  moved 
from  P.  This  direction  may  be  indicated  in  various  ways ;  by 
saying,  for  instance,  that  the  point  moves  toward  the  right  from 
P,  or  toward  the  left  from  P;  that  the  point  moves  toward  N,  or 
toward  M;  that  the  point  moves  in  the  direction  of  JSf,  or  in  the 
direction  of  M;  and  so  on.     Mathematicians,  engineers,  and  others 

F  P       B  CD 

I I       I I       I 

M  N 

riG.  31. 

have  agreed  to  use  a  particular  method  (and  this  practically  comes 
to  the  adoption  of  a  particular  rule)  for  indicating  the  two  oppo- 
site directions  in  which  a  point  can  move  along  a  line,  or  in  which 
distances  along  a  line  can  be  measured.  This  convention,  or  rule 
which  has  been  adopted  for  the  sake  of  convenience,  is  as  follows : 

Distances  measured  along  a  line,  or  along  parallel  lines,  in  one 
direction  shall  he  called  positive  distances,  and  shall  be  denoted  by 
the  sign  +/  distances  measured  iyi  the  opposite  direction  shall  be 
called  negative  distances,  and  shall  be  denoted  by  the  sign  — . 

The  convenience  of  this  custom,  fashion,  or  rule,  will  become 
apparent  in  the  examples  that  follow.*     In  Fig.  31  let  distances 

*  Advances  in  mathematics  have  often  depended  upon  the  introduction  of 
a  good  custom  which  has  at  last  been  universally  adopted  and  made  a  rule. 
Thus,  for  example,  the  custom  of  using  exponents  to  show  the  power  to 
which  a  quantity  is  raised,  which  was  first  introduced  in  the  first  half  of  the 
sixteenth  century,  and  made  gradual  progress  until  its  final  establishment  in 
the  latter  half  of  the  seventeenth  century,  has  beeu  of  great  service  in  aiding 
the  advances  of  algebra. 

65 


66  PLANE  TRIGONOMETRY,  [Ch.  V, 

measured  in  the  direction  of  JVbe  t2^Q\i  positively ;  then  distances 
measured  in  the  direction  of  M  will  be  taken  negatively.  On 
directed  lines  the  direction  in  which  a  line  is  measured,  or  in 
which  a  point  moves  on  a  line,  is  indicated  by  the  order  of  the 
letters  naming  the  line.  Thus,  for  example,  if  a  point  moves  from 
B  to  (7,  the  distance  passed  over  is  read  BC.  In  this  reading,  the 
starting  point  is  indicated  by  the  first  letter  B,  and  the  stopping 
point,  by  the  last  letter  C.  After  the  same  fashion,  CB  means  the 
distance  from  C  to  B.  If,  for  instance,  there  are  3  imits  of  length 
between  B  and  (7,  then  ^(7  =  +  3,  05  =  -  3. 

EXAMPLES. 

1.  Suppose  a  point  (Fig.  31)  moves  from  P  to  J5,  thence  to  C,  thence  to 
Z>,  thence  to  F.  Let  the  number  of  units  of  length  between  P  and  i?,  B  and 
O,  C  and  Z>,  F  and  D,  be  2,  3,  2,  10,  respectively.  The  point  starts  at  P  and 
stops  at  F  ;  hence  the  distance  from  the  starting  point  to  the  stopping  point 
is  PF.  In  this  case  the  point's  trip  from  P  to  i^  is  made  in  several  steps  as 
indicated  above.    That  is,  on  properly  indicating  the  lines  passed  over, 

PF=PB-hBC  +  CD  +  DF 

=  2  +  3  +  2-10     [•.•  FD  =  +  10,  then  DF  =  -  10.] 
=  -3. 

This  shows  that  the  final  position  of  the  moving  point  is  three  units  to  the 
left  of  P.  This  example  also  shows  one  great  convenience  of  the  rule  of 
signs  in  measurement^  namely,  that  by  attending  to  this  rule  and  to  the 
proper  naming  of  the  lines  passed  over  by  a  moving  point,  one  immediately 
obtains  the  result  of  the  successive  movements.    ' 

Note.  In  the  following  examples,  in  lines  that  lie  east  and  west,  let 
measurements  toward  the  east  be  taken  positively ;  in  lines  that  lie  north 
and  south,  let  measurements  toward  the  north  be  taken  positively. 

2.  A  man  travelling  on  an  east  and  west  line  goes  east  20  mi.,  then  east 
16  mi.,  then  west  18  mi.,  then  east  30  mi.  What  is  his  final  distance 
from  the  starting  point  ?  [Draw  a  figure,  and  indicate  the  successive  trips 
by  letters.] 

3.  A  man  travelling  on  an  east  and  west  line  goes  west  20  mi.,  then  east 
10  mi.,  then  east  25  mi.,  then  east  30  mi.,  then  west  45  mi.     Do  as  in  Ex.  2. 

4.  A  man  travelling  on  a  north  and  south  line  goes  north  100  mi.,  then 
south  60  mi.,  then  south  110  mi.,  then  north  200  mi.,  then  north  15  mi., 
then  south  247  mi.     Do  as  in  Ex.  2. 


37.]  TRIGONOMETBIC  RATIOS   OF  ANGLES.  67 

37.  Trigonometric  definition  of  an  angle.  Angles  unlimited  in 
magnitude.  Positive  and  negative  angles.  In  books  on  plane 
geometry  a  plane  angle  is  defined  in  various  ways,  namely,  as  the 
inclination  of  two  lines  to  one  another,  which  meet  together,  but 
are  not  in  the  same  direction;  or,  as  the  figure  formed  by  two 
straight  lines  drawn  from  the  same  point;  or,  as  the  amount  of 
divergence  of  two  lines  which  meet  in  a  point,  or  would  meet  if 
produced;  or,  as  the  opening  between  two  straight  lines  which 
meet ;  or,  as  the  difference  in  direction  of  two  lines  which  meet ; 
and  so  on.  In  these  definitions  an  angle  is  always  regarded  as 
less  than  two  right  angles.  A  definition  according  to  which 
angles  are  less  restricted,  is  adopted  in  trigonometry. 

Trigonometric  definition  of  an  angle.  The  angle  between  two 
lines  which  intersect  is  the  amount  of  turning  which  a  line 
revolving  about  their  point  of  intersection  makes,  when  it  begins 
its  revolution  at  the  position  of  one  of  the  two  lines  and  stops  in 
the  position  of  the  other  line.  Thus,  for  example,  the  angle 
between  OX  and  OQ  is  the  amount  of 
turning  which  is  made  by  a  line  OP 
revolving  about  0  when  OP  starts  re- 
volving from  the  position  OX  and  stops 
its  revolution  at  the  position  OQ.  The 
line  OX  at  which  the  revolution  begins, 
is  called  the  initial  line;  the  line  OQ  at 
which  the  revolution  ends,  is  called  the 
terminal  line ;  when  the  turning  line  OP 
has  reached  the  terminal  position  OQ, 
OP  is  said  to  have  described  the  angle 
XOQ. 

Let  YOTi  be  at  right  angles  to  XiOX.  When  OP  has  revolved 
until  it  lies  in  the  position  0  Y,  it  has  described  a  right  angle,  or 
90° ;  when  it  has  revolved  until  it  lies  in  the  position  OXi,  it  has 
described  two  right  angles,  or  180°  (this  is  usually  termed  "a 
straight  angle  "  or  "  a  flat  angle '') ;  when  OP  keeps  on  turning 
until  it  is  in  the  position  0  Fj,  it  has  described  three  right  angles, 
or  270°;  when  OP  has  again  reached  the  position  OX,  that  is, 
when  it  has  made  one  complete  revolution,  it  has  described  four 
right  angles,  or  360°. 


68 


PLANE  TRTGONOMETBT, 


[Ch.  V. 


Angles  unlimited  in  magnitude.  Now  OP  may  start  revolving 
from  OX,  make  one  complete  revolution,  continue  to  revolve, 
and  then  cease  revolving  when  it  has  again  reached  the  position 
OQ.  This  is  indicated  in  Fig.  33.  Or, 
OP  may  make  two  complete  revolutions 
before  it  comes  to  rest  in  the  position  OQ ; 
or,  it  may  make  three  revolutions,  or  four, 
or  as  many  as  one  please,  before  ceasing 
its  revolution  at  the  position  OQ.  An  an- 
gle of  360°  is  described  each  time  that  OP 
makes  a  complete  revolution,  and  OP  can  make  as  many  revolu- 
tions as  one  please.  According  to  the  trigonometric  definition 
of  an  angle,  therefore,  angles  are  unlimited  in  magnitude. 

Moreover,  when  this  definition  of  an  angle  is  adopted,  the  same 
figure  can  represent  an  infinite  number  of  different  angles*     Any  two 
of  these  angles  differ  from  each  other  by  a 
whole  number  of  complete  revolutions.     For 
instance,  Fig  34  may  represent  60°,  360° -f  60° 
or  420°,  2  .  360°  +  60°  or  780°,  3  •  360°  +  60° 
or  1140°,  '",n-  360°  +  60°,  in  which  n  denotes 
any  whole  number.     A7iy  ttvo  of  these  angles     i 
differ  by  a  multi2:)le  o/360°.    Angles  which  have 
the  same  initial  and  terminal  lines  may  be 
called  coterminal  angles. 

Positive  and  negative  angles.  The  revolving  line  OP  (Fig.  32) 
may  revolve  about  0  in  the  same  direction  as  that  in  which  the 
hands  of  a  watch  revolve,  or  it  may  revolve  in  the  opposite  direc- 
tion. The  following  convention  (see  Art.  36)  has  been  adopted 
for  the  sake  of  distinguishing  these  two  opposite  directions : 

When  the  turning  line  revolves  in  a  counter-clockwise  direction, 
the  angles  described  are  said  to  be  positive,  and  are  given  the  plus 
sign;  when  the  turning  line  revolves  in  a  clockwise  direction,  the 
angles  described  are  said  to  be  negative,  and  are  giveii  the  minus  sign. 
Thus,  for  example.  Fig.  34  represents  the  angles  -f  60°,  —  300°; 
further,  this  figure  represents  the  angles  60°  ±  7i  •  360°,  in  which  n 
denotes  any  whole  number.  The  angle  —300°  is  included  in 
these  angles,  for,  on  putting  —1  for  n,  there  is  obtained  60°— 360°, 
i.e.  —  300°.     (Negative  angles  are  also  unlimited  in  magnitude.) 


Initial  Line 


Fia.  34. 


38.]         SUPPLEMENT  AND   COMPLEMENT.  69 

As  in  the  case  of  lines,  the  sign  of  an  angle  can  he  denoted  by  the 
order  of  the  letters  used  in  naming  the  angle.  Thus  XOQ  denotes 
the  angle  formed  by  revolving  OX  toward  OQ,  and  QOX  denotes 
the  angle  formed   by  revolving  OQ  toward  OX.     Accordingly, 

qox=  -XOQ. 

Quadrants.     In  Fig.    32,    XOY,   YOX^,    X^OT^,   Y,OX,    are 

called  the  first,  second,  third,  and  fourth  quadrants,  respectively. 
When  the  turning  line  ceases  its  revolution  at  some  position  be- 
tween OX  and  0  Y,  the  angle  described  is  said  to  be  an  angle  in 
the  first  quadrant;  when  the  final  position  of  the  turning  line  is 
between  0  Y  and  OXj,  the  angle  described  is  said  to  be  in  the  sec- 
ond quadrant;  and  so  on  for  the  third  and  fourth  quadrants. 

For  example,  the  angles  30°,  -  345°,  395°,  725°  are  all  in  the  first  quad- 
rant ;  the  angles  —  60°,  340°,  710°  are  all  in  the  fourth  quadrant ;  the  angle 
—  225°  is  in  the  second  quadrant,  and  the  angle  225°  is  in  the  third  quadrant. 

Note.  While  all  acute  angles  are  in  the  first  quadrant,  all  angles  which 
are  in  the  first  quadrant  are  not  acute. 

EXAMPLES. 

Note.  When  it  is  necessary,  the  number  of  revolutions  and  their  direction 
may  be  indicated  on  the  figure  in  the  manner  shown  in  Fig.  34. 

Lay  off  the  following  angles  with  the  protractor:  In  the  case  of  each 
angle  name  the  least  positive  angle  that  has  the  same  terminal  line.  Name 
the  quadrants  in  which  the  angles  are  situated.  In  the  case  of  each  angle 
name  the  four  smallest  positive  angles  that  have  the  same  terminal  line. 

1.  137°,  785°,  321°,  930°,  840°,  1060°,  1720°,  543°,  3657°. 

2.  _  240°,  -  337°,  -  967°,  -  830°,  -  750°,  -  1050°,  -  7283°. 

3.  _  470  ^  230°  +  37°,  420°  -  470°  +  210°  -  150°,  230°  -  47°  +  37°,  230° 
+  37°  -  47°. 

38.  Supplement  and  complement  of  an  angle.  TJie  supplement  of 
an  angle  is  that  angle  which  must  be  added  to  it  in  order  to  make 
two  right  angles,  or  180°;  the  complement  of  an  angle  is  that 
angle  which  must  be  added  to  it  in  order  to  make  one  right 
angle,  or  90°.     Thus,  if  A  be  any  angle,  then 

supplement  of  angle  A  =  180°  —  A, 

complement  of  angle  A  =    90°  —  A. 


70  PLANE  TRIGONOMETRY,  [Ch.  V. 

EXAMPLES. 

1.  What  are  the  complements  and  supplements  of  40°,  227°,  —  40°  ? 

complement  of    40°=    90°-    40°=    50°; 

supplement  of    40°  =  180°  -    40°  =  140°. 

complement  of  227°  =    90°  -  227°  =  -  137°  ; 

supplement  of  227°  =  180°  -  227°  =  -  47°. 

complement  of  -  40°  =    90°  -  (-  40°)  =  130°  ; 

supplement  of  -  40°  =  180°  -  (-  40°)  =  220°. 

2.  By  means  of  a  figure  verify  the  results  obtained  in  Ex.  1. 

3.  What  are  the  complements  of  -  230°,  150°,  -  40°,  340°,  75°,  83°,  12°, 
-  295°,  -  324°,  200°,  240°,  -  110°,  -  167°  ? 

4.  What  are  the  supplements  of  the  angles  in  Ex.  3  ? 
6.   Verify  the  results  in  (3),  (4),  by  drawing  figures. 

39.   The  convention  of  signs  on  a  plane.     Articles  36,  37  contain 
statements  of  the   conventions  adopted   regarding  the  algebraic 

signs  to  be  given  to  distances  measured 
on    parallel    straight   lines,   and    to 
angles  described  by  the  revolution  of 
a  turning  line.   A  figure,  such  as  Figs. 
_^__      32,  35,  will  be  frequently  used  in  the 
j  -^     articles  that  follow.     In  this  figure, 
!  OX  is  the  initial  line,  the   turning 

P4*  line  revolves  about  O,  and  YOY^  is 

at  right  angles  to  X^OX.     The  fol- 
PiG.  35.  lowing  convention  has  been  adopted  re- 

garding the  lilies  which  will  be  used : 

Horizontal  lines  measured  in  the  direction  of  X  are  taken  positively  ; 
Horizontal  lines  measured  in  the  direction  of  Xi  are  taken  negatively  ; 
Vertical  lines  measured  upward  are  taken  positively  ; 
Vertical  lines  measured  downward  are  taken  negatively. 

The  distance  of  points,  such  as  Pj,  P2,  P3,  Pi,  from  X^X,  is 
always  measured /rom  X^X  toward  the  points. 

Any  turning  line  (or  oblique  line)  as  OP  is  measured  positively 


-P. 


X, 


40.] 


GENERAL  DEFINITION   OF  RATIOS. 


71 


from  0  toward  the  end  of  the  turning  line  which  lies  in  the 
direction  of  X  from  O  when  the  turning  line  coincides  with  the 
initial  line.  Thus  a  distance  +  3  on  OP  will  terminate  at  T,  dis- 
tant 3  units  from  0,  and  a  distance  —  3  on  OP  will  terminate  at 
Ti,  distant  3  units  from  0,  but  in  the  direction  opposite  to  the 
former.  This  is  sometimes  briefly  expressed  in  the  words:  the 
turning  line  carries  its  positive  direction  with  it  in  its  revolution, 

40.  General  definition  of  the  trigonometric  ratios.  The  remarks 
in  this  article  apply  to  each  of  the  four  figures  below.  In  each 
figure,  O  is  the  point  about  which  the  angle  is  described,  OX  is 
the  initial  line,  and  OP  is  the  terminal  line.  The  first  figure 
represents  any  angle  in  the  first  quadrant;  the  second  figure 
represents  any  angle  in  the  second  quadrant ;  the  third  figure,  any 
angle  in  the  third  quadrant ;  and  the  fourth  figure,  any  angle  m 
the  fourth  quadrant.     In  each  figure  the  angle  will  be  called  A. 


X 


M  X 


Fig.  36. 


Let  P  be  any  point  in  OP,  the  terminal  line  of  any  angle  A. 
From  P  draw  PM  at  right  angles  to  the  initial  line  OX,  or  to  the 
initial  line  produced  in  the  negative  direction.  In  each  figure, 
OM  is  the  distance  measured  along  X^OX  from  the  point  0  to 
the  foot  of  the  perpendicular  MP,  and  MP  is  the  distance  from 
XiOX  to  the  point  P.  Following  are  the  definitions  of  the  trigo- 
nometric ratios ;  these  definitions  apply  to  the  angles  represented 
in  Fig.  36,  and,  accordingly,  to  all  angles  whatsoever.  ^Particular 
attention  should  be  j^aid  to  the  order  of  the  letters  used  in  naming 
the  lines,  for  this  order  iiidicates  the  direction  in  which  the  line  is 
measured.     See  Art.  36.]   * 


72  PLANE  TBIGONOMETRT,  [Ch.  V. 

MP 

The  ratio  -— ■  is  called  the  sine  of  the  angle  A, 

The  ratio  -— — -  is  called  the  cosiyie  of  the  angle  A. 

MP 

The  ratio  -— — :  is  called  the  tangent  of  the  angle  A, 

The  ratio  •— —-  is  called  the  cotangent  of  the  angle  A. 
MP 

OP 

The  ratio  -- —  is  called  the  secant  of  the  ansrle  A, 
OM 

OP 
The  ratio  -— —  is  called  the  cosecant  of  the  angle  A, 
MP 


These  definitions  may  be  briefly  stated : 

(1) 


'•»-=^-    — ^-     ^--W,- 


A  =  H~'         cot  A  =  ^^»         cosec  A  = 


OP  MP  MP 

Inspection  will  show  that  the  definitions  of  the  trigonometric 
ratios  for  acute  angles  given  in  Art.  12,  are  in  accordance  with 
these  general  definitions. 

N.B,  The  projection  definitions  of  the  trigonometric  ratios  are  given  in 
Note  13,  Appendix. 


41.  The  algebraic  signs  of  the  trigonometric  ratios  for  angles 
in  the  different  quadrants.  Figures  36  show  that  if  the  angle  A 
is  in  the  first,  second,  third,  fourth  quadrants,  then  the  algebraic 
sign  of  MP  is  +,  +,  — ,  — ,  respectively,  and  the  algebraic  sign 
of  OM  is  +,  ^,  — ,  4-,  respectively.  As  stated  in  Art.  39, 
OP  is  always  taken  positively.  Hence,  on  paying  regard  to  the 
algebraic  signs  of  OM,  MP^  OP,  in  the  several  quadrants,  it  will 
be  seen  that  the  ratios  of  the  angles  in  these  quadrants  are  posi- 
tive or  negative,  as  indicated  in  the  fc^lowing  table : 


41.] 


ALGEBRAIC  SIGNS   OF  RATIOS. 


73 


Quadrant, 

I. 

II. 

III. 

IV. 

Sine 

+ 

+ 





Cosine 

+ 

— . 

— 

+ 

Tangent          .... 

+ 

— 

+ 

- 

Cotangent      .... 

-f 

— 

+ 

— 

Secant    .        .    *    . 

-1- 

— 

— 

+ 

Cosecant        .... 

+ 

+ 

— 

— 

Fig.  37. 


EXAMPLES. 

The  student  is  advised  to  preserve  his  work  on  these  examples.  If  he 
regards  his  results  attentively,  he  will  probably  discover  some  useful  facts, 
and  he  able  to  deduce  some  useful  theorems^  concerning  angles  in  general. 
Any  preceding  results,  such  as  those  in  Art.  15,  may  be  used  as  an  aid  in 
solving  these  exercises. 

1.  Find  the  ratios  of  945°.  -^i  ^  "1     ^ ^ 

945°  =  2  X  360°  +  225°. 

.••  OP,  the  terminal  line  of  angle  945°,  has  the 
position  shown  in  Fig.  37.  For  this  position  of 
the  terminal  line,  03/ and  ilfP  are  both  negative. 
As  shown  in  Art.  15,  the  lines  OM,  MP,  OP, 

in  this  figure  are  respectively  proportional  to  1,  1,  y/2.  These  are  indicated 
on  the  figure  with  their  proper  algebraic  signs.  It  is  immediately  apparent 
that 

sin  945°=--^,    cos  945°  =  - -i^,    tan  945°  =+1,    cot  945°  =+1, 

\/2  \/2 

sec  945°  =  -  \/2,     cosec  945°  =  -  \/2. 

2.  Construct,  and  find  the  ratios  of,  420°,  780°,  1140°. 

3.  Construct,  and  find  the  ratios  of,  120°,  480°,  240°,  600°,  -  60°,  300°, 
660°,   -720°. 

4.  Construct,  and  find  the  ratios  of,  150°,  410°,  210°,  -150°,  330°,  -390°. 

5.  Construct,  and  find  the  ratios  of,  45°,  765°,  135°,  -  225°,  225°,  585°, 
-405°,  1035°. 

6.  Construct,  and  find  the  ratios  of,  -  754°,  487°,  -  245°. 

7.  Compare  the  ratios  of  90°  -  30°,  90°  -  60°,  90°  -  45°,  90°  -  135°, 
90°  -  240°,  90°  -  300°,  with  the  ratios  of  30°,  60°,  45°,  135°,  240°,  300°, 
respectively. 


74 


PLANE  TRIGONOMETRY, 


[Ch.  V. 


8.  Compare  the  ratios  of  90°  +  30°,  90°  +  60°,  90°  +  45°,  90°  +  135°, 
90°  +  240°,  90°  +  300°,  with  the  ratios  of  30°,  60°,  45°,  135°,  240°,  300°, 
respectively. 

9.  Compare  the  ratios  of  180°  -  30°,  180°  -  60°,  180°  -  45°,  180°  -  135°, 
180°  -  240°,  180°  -  300°,  with  the  ratios  of  30°,  60°,  45°,  135°,  240°,  300°, 
respectively.     So,  also,  the  ratios  of  —  30°,  —  60°,  —  45°,  etc. 

10.  Are  any  general  relations  indicated  by  the  results  of  Exs.  7,  8,  9  ? 
If  so,  state  these  relations.     Try  to  prove  them. 

42.  To  represent  the  angles  geometrically  when  the  ratios  are 
given.  In  constructing  the  angles  in  tins  article  it  is  necessary 
to  bear  in  mind  that,  according  to  the  definitions  given  in  Art.  40  : 

When  MP  is  positive,  it  can  he  drawn  in  the  Jirst  and  second  quadrants; 
When  MP  is  negative,  it  can  be  drawn  in  the  third  and  fourth  quadrants; 
When  OM  IS  positive,  it  is  to  be  drawn  in  the  direction  OX; 
When  OM  is  negative,  it  is  to  be  drawn  in  the  direction  OXi ; 
and      OP  is  to  be  taken  positively. 


EXAMPLES. 

1.   Represent  by  a  figure  the  angles  which  have  sines  equal  to  f .    Calcu- 
late their  other  ratios.     Let  A  denote  an  angle  whose  sine  is  | ;  i.e.  let 

sin^  =  f.     But  sin ^  =  :^  (Art.  40).     Hence,  iJfP:OP  =  3:4;   and   if 

OP  =  4,  then  MP  =  3.  Now,  MP  can  be  drawn .  positively  in  both  the  first 
and  second  quadrants.  Hence  the  problem  amounts  to  finding  a  point  in 
the  first  quadrant  and  a  point  in  the  second  quadrant,  each  at  a  distance  4 

from  0  and  a  distance  3  from  XiOX.  The 
result  is  indicated  in  Fig.  38.  The  student 
can  make  the  construction  for  himself. 
The  angles  having  sines  equal  to  f,  accord- 
ingly, include  all  the  angles  which  have  OP 
for  a  terminal  line,  and  all  the  angles  which 
have  OPi  for  a  terminal  line.  By  Art.  37 
each  of  these  two  sets  of  angles  consists  of 
an  infinite  number  of  angles,  any  two  of 
which  differ  from  one  another  by  a  whole 
number,  positive  or  negative,  of  complete 
revolutions.  A  general  algebraic  expression  which  includes  all  these  angles, 
iti  deduced  in  Art.  85. 


X,   M, 


42-43.]  GEOMETRIC  REPRESENTATIOK.  75 

Figure  38  shows  that  for  angles  having  OP  for  a  terminal  line,  cosine  is 

■ — ,  tangent  is  — -,  etc.  ;  and  that  for  angles  having  OPi  for  a  terminal  line, 

4  _  V7 

^-7  Q 

cosine  is ,  tangent  is  — — ,  etc.      Since  the  given  sine  is  positive,  it  is 

4  VV 

apparent  that  the  angles  required,  must  be  in  the  first  and  second  quadrants. 
(See  Art.  41.) 

2.  Bepresent  by  a  figure  all  the  angles  which  have  tangents  equal  to 
—  f .     Let  A  denote  an  angle  whose  tangent  is  —  | ;   i.e.  let  tan  A--^. 

prhis  may  be  written,  i-^  or  — -1     But  tan  A  =  ^^^  (Art.  40).     Hence, 

if  MP  =  3,  then  OM  =  -i,  and  if  MP  =  -  3,  then  031  =  4.     When  3IP  is 
positive  and  03/ is  negative,  OP  can  lie  p-v 

only  in  the  second  quadrant.     When  1  ^v 

MP  is  negative  and  OM  is  positive,  3|         >s. 

OP  can  lie  only  in  the  fourth  quadrant.  j      _      _    ^s^  ilfi 

Figure  39  represents  the  angles.     The     x    M        -4 
student  can  make  the  construction  for 
himself.     Thus,  the  angles  whose  tan- 
gents are  equal  to  —  f ,  consist  of  the 
set  of  angles,  infinite  in  number,  which  ^^^'  ^' 

have  OP  for  a  terminal  line,  and  the  set  of  angles,  infinite  in  number,  which 
have  OPi  for  a  terminal  line. 

3.  Calculate  the  other  ratios  of  the  angles  in  Ex.  2. 

4.  Represent  geometrically  all  the  angles  whose  cosine  is  |.    Calculate 
their  other  ratios. 

5.  So,  also,  when  the  cosine  is  —  |. 

r  4         14        4-1 

6.  So,  also,  when  the  tangent  is  |.  Note.     -  =  -^—~  =  — -• 

7.  So,  also,  when  the  sine  is  —  f . 

8.  So,  also,  when  the  secant  is  f. 

9.  So,  also,  when  the  secant  is  —  |. 

10.   So,  also,  when  the  cosecant  is  —  2  ;  |. 

N.B.  The  student  is  now  strongly  recommended  to  delay  the  reading  of 
the  next  article  until  after  he  has  reviewed  the  properties  stated  in  Art.  13, 
and,  if  possible,  determined  what  are  the  correct  corresponding  statements  for 
angles  in  general. 

43.   Connection  between  angles  and  the  trigonometric  ratios.     For 

the  same  terminal  position  of  the  revolving  line  OP  each  of  the 

MP 

ratios,  — — ,  etc.,  in  (1)  Art.  40,  is  always  the  same,  no  matter 


76  PLANE  TRIGONOMETRY,  [Ch.  V. 

where  the  point  P  is  taken  on  the  revolving  line.     Thus,  for 
example,  let  any  other  point  Pi  (Fig.  40  a)  be  taken  on  the  ter- 


X^  M  Ml       O  X  X^M^M  O  X 

Fig.  40a.  Fig.  40b. 

minal  line,  and  let  PiMi  be  drawn  perpendicular  to  X^OX.     Thtv. 

^—-  —  ^   \     That  is,  the  sine  of  any  angle  whose  terminal  line 
OP       OPi 

is  OP,  has  a  fixed  definite  value.     The  same  can  be  shown  for 

the  other  ratios.     Hence,  as  already  shown  in  Art.  13  for  angles 

between  0°  and  90^ 

(1)  To  each  angle  there  corresponds  but  one  value  of  each  trigono- 
metric ratio. 

Now  let  OP  revolve  a  little  from  OP  into  the  position  OPi 
(Fig.  40  b).  For  convenience  keep  OPi  equal  to  OP.  Draw 
PiMi  at  right  angles  to  X^OX,     Then, 

MP 

the  sine  of  the  angle  whose  terminal  line  is  OP  —  — — , 

MP 

and  the  sine  of  the  angle  whose  terminal  line  is  OPi  =  — ^— ^» 

OPi 

Since  MiP^  is  not  equal  to  MP,  it  follows  that  these  two  sines 
are  unequal.  Hence,  the  sine  of  an  angle  changes  when  the  angle 
changes.     The  same  can  be  shown  for  the  other  ratios.     Hence, 

(2)  TJie  ratios  of  an  angle  change  ivhen  the  angle  changes. 

The  variation  in  the  ratios  as  the  angle  increases,  is  discussed  in  Art.  77. 

Ex.  1.  In  the  above,  OPi  is  taken  equal  to  OP.  Why  does  this  not  affect 
the  generality  of  the  deduction  ? 

Ex.  2.  Trace  the  changes  in  the  trigonometric  ratios  as  the  turning  line 
revolves  from  0°  to  360°.  Compare  your  results  with  those  of  Art.  15  (7,  and 
those  given  in  the  table  at  the  end  of  Art.  77. 

It  has  been  shown  in  Art.  37  that  an  infinite  number  of  angles 
have  the  same  terminal  line.     Jt  follows  that  in  each  of  the 


44.]  RELATIONS   OF  TRIGONOMETRIC  RATIOS.  11 

MP 

figures  in  Art.  40,  -——  is  tlie  sine  of  an  infinite  number  of  angles. 

The  same  is  true  in  the  case  of  the  other  ratios.  Moreover,  the 
geometrical  solutions  in  Art.  42  show  that  there  are  two  sets  of 
angles  correspo7iding  to  each  given  ratio,  and  that  each  set  is  infinite 
in  number  J  and  has  a  particular  terminal  line.     Hence, 

(3)  To  each  value  of  a  trigonometric  ratio  there  corresponds  an 
infinite  number  of  angles. 

Note.  The  student  will  see,  by  turning  to  Arts.  84-87,  that  all  angles 
which  have  the  same  sine,  can  he  given  in  a  simple  formula,  and  that  the 
same  fact  is  true  in  the  case  of  each  of  the  other  ratios.  The  deduction  of 
these  formulas,  while  easily  possible  at  this  place,  is  postponed  in  order  to 
permit  the  early  completion  of  the  solution  of  triangles. 

44.  Relations  between  the  trigonometric  ratios  of  an  angle.     The 

relations  between  the  trigonometric  ratios  of  an  acute  angle  were 
set  forth  in  Art.  18.  It  will  now  be  shown  that  these  relations 
also  hold  for  the  ratios  of  any  angle. 

A.  Inspection  of  the  definitions  (1),  Art.  40,  shows  the  reciprocal 
relations,  namely : 


sin ^ cosec ^ -  1 ;    cos^sec^  =  l;    tan ^ cot 4  =  1. 

(1) 

B.  In  each  of  the  figures  in  Art.  40, 

MP                                          OM 
t,r,  1-^P-OP      slnA.    ^^.   .OM       OP      ^A 
*^'*      OM      OM     cos  A'    *"*'*      MJP     MP     sin  a' 

OP                                             OP 

(2) 

C.  In  each  of  the  figures  in  Art.  40, 

On  dividing  both  members  of  this  equation  by  OP  ,  OM  ,  MP  , 
in  turn,  and  following  the  same  process  as  that  adopted  in  Art.  18, 
it  results  that 

sm^A  +  cos^A  =  l;  sec^^  =  1 +tan2^;  cosec^ JL  =  1  + cot^ ^.    (3) 

From  the  first  of  relations  (3)  it  follows  that 
cos  -4  =  ±  Vl  —  sin^  A, 


78  PLANE   TRIGONOMETRY.  ICu.  V. 

This  shows  that,  corresponding  to  a  given  sine,  there  are  two 
cosines  which  are  numerically  equal,  and  opposite  in  algebraic 
sign.  Ex.  1,  Art.  42,  illustrates  this.  This  is  also  manifest  in 
the  table  of  signs  in  Art.  41.  As  indicated  in  this  table,  the  sine 
is  positive  in  the  first  and  second  quadrants,  and  then  the  cosine 
is  positive  and  negative,  respectively ;  the  sine  is  negative  in  the 
third  and  fourth  quadrants,  and  then  the  cosine  is  negative  and 
positive,  respectively.  The  signs  of  the  remaining  ratios  corre- 
sponding to  a  given  sine  will  be  apparent  on  a  short  geometrical 
inspection,  or  by  a  glance  at  this  table  of  signs.  When  any  single 
ratio  is  given,  there  is  an  ambiguity  as  to  the  signs  of  some  of  the 
other  ratios.  Thus,  to  take  another  instances-it  follows  from 
the  second  of  (3)  that 


I 


tan  A  =  ±  Vsec^  A  —  1. 

The  secant  of  A  is  positive  in  the  first  and  fourth  quadrants^ 
and  then  the  tangent  is  positive  and  negative  respectively;  the 
secant  of  A  is  negative  in  the  second  and  third  quadrants,  and 
then  the  tangent  is  negative  and  positive  respectively.  The 
double  sign  which  appears  in  these  relations  was  referred  to  in 
the  examples.  Art.  18.  The  student  is  advised  to  review  and 
work  the  examples,  for  angles  in  general,  in  Art.  18. 


EXAMPLES. 

1.    Given  that  sin  ^  =  | ;  find  the  other  ratios  of  A  by  means  of  the 
relations  shown  in  this  article. 

[In  Ex.  1,  Art.  42,  this  problem  is  solved  geometrically ;  here  it  will  be 
solved  algebraically. '\ 

±V7.     ...  ._     1     _     4     . 


cos  ^  =  ±  V 1  —  sin"'^  A  =  — ;     sec  A 


4  cos  A      ^  ^^7 

cosec^=-J-=i;    tanyl  =  ^iil^  =  -J-;     cot  A  = -^— =  ±2^. 
sin. A     3  cos^      ±Vl  tan  A         3 

Since  the  given  sine  is  positive,  the  corresponding  angles  are  in  the  first 
and  second  quadrants.  Hence  the  double  values  of  the  calculated  ratios  are 
paired  as  follows : 


46.]  BATIOS  OF  90'' -  A,   ETC.  79 

sin  A         cos  A         tan  A         cot  A         sec  A        cosec  ^ 


3 

+  V7 

3 

V7 

4 

4 

4 

4 

V7 

3 

V7 

3 

3 

-V7 

3 

-V7 

4 

4 

4 

4 

V7 

3 

V7 

3 

Find  the  other  ratios  algebraically ,  and  verify  the  results  geometrically^ 
when : 

2.  cos  ^  =  —  |.        3.  tan  ^  =  f .        4.  sec  A  =  A.  5.  cosec  A  =  —  b. 

6.  sin  ^  =  —  f        7.  cos  A  =  ^.        8.  tan  J.  =  —  3.       9.      cot  A  =  l. 

Find  the  other  ratios  algebraically,  and  verify  the  results  geometrically, 
when  angle  J.  satisfies  the  following  p«iVs  o/conrfiYiows:  .   / 

10.   sin  ^  =  i     and  tan  A  = 11.   tan  ^1  =  VS  and  sec  ^  =  —  2. 

V3 

\/5 
12.   cos  J.  =  —  I  and  sin  ^  =  +  — —        13.  sin  ^  =  —  f  and  tan  A  =  \, 

o 

Give  a  geometrical^lution  of  the  following  trigonometric  equations  : 

14.   sin  A  =  l,    "   15.   cos  ^  =  -  f        16.   tan  ^  =  4.         17.   sec  A  =  5. 

Namoi^^Rur  least  angles,  and  also  the  four  least  positive  angles,  that 
satu^PPI^quations : 

18.  sin^=— .       19.  tan^=\/3.       20.  cos^= — —»      21.  cot^=-\/3. 
\/2  V2 

45.  Ratios  of  90°  -  A,  180°  -  A,  90°  -{-A,  -A,  compared  with 
the  ratios  of  Ay  A  being  any  angle.  The  student  may  have 
suspected,  from  Exs.  7-10,  Art.  41,  that  there  is  a  close  connec- 
tion between  the  ratios  of  an  angle  A  on  the  one  hand,  and  the 
ratios  of  the  angle  —  A  and  of  angles  differing  from  A  and  —  A 
by  multiples  of  90°,  on  the  other.  He  may  have  discovered 
already  what  the  connection  is.  This  connection,  which  is  set 
forth  in  this  article,  is  interesting  in  the  study  of  angles,  and 
has  an  important  bearing  on  the  construction  of  trigonometric 
tables,  and  on  the  solution  of  triangles. 

In  each  figure  in  this  article  OP  is  the  terminal  line  of  the 
angle  A,  and  OP^  is  the  terminal  line  of  the  related  angle  which 
is  under  consideration ;  for  the  purpose  of  easy  comparison,  OPi 
is  always  taken  equal  to  OP]  MP,  M^P^,  are  the  perpendiculars 
drawn  from  the  initial  line  to  P,  Pj,  respectively.    The  deductions 


80 


PLANE  TRIGONOMETRY, 


[Ch.  V. 


made  in  the  simplest  case,  namely,  when  A  is  an  angle  in  the  first 
quadrant,  are  true  for  all  angles.  The  student  is  advised  to  con- 
sider only  the  simplest  case,  when  first  he  considers  the  subject  of 
this  article,  and  then  to  draw  the  figures  and  make  the  deduc- 
tions for  himself,  in  the  cases  in  which  angle  A  is  in  the  second, 
third,  and  fourth  quadrants,  respectively. 

Note.  A  compound  angle,  90°  —  A^  for  instance,  can  be  described  by- 
revolving  the  turning  line  forward  through  90'',  and  then  backward 
through  an  angle  equal  to  A;  or,  these  steps  may  be  taken  in  a  reverse 
order,  namely,  by  revolving  the  turning  line  backward  through  an  angle 
pqual  to  A^  and  then  forward  through  90°.  Similarly,  for  the  compound 
yr;l<^s  90°  +  J,  180°  ±  A,  etc. 

For  the  sake  of  clearness  of  construction,  it  is  better  not  to  take  the 
terminal  line  of  A  nearly  midway  between  XiOXand  Y\OY. 

A.  Ratios  of  90°  -  A.  Describe  the  angles  A,  90°  —  A.  Let  OP, 
OPi,  be  the  terminal  lines  of  A,  90°  —  A,  respectively.     In  Figs. 


FiQ.  41a. 


Fig.  41t). 


Fig.  flOr 


Fig.  41d. 


41  a,  41  b,  41  c,  41  d,  A  is  an  angle  in  the  first,  second,  third,  and 
fourth  quadrants,  respectively. 

Take  OP^  equal  to  OP,  and  draw  MP,  M^P^,  at  right  angles  to 
the  initial  line.  In  the  triangles  M^OPi,  MOP,  (in  each  figure)  the 
angles  at  M^,  M,  are  right  angles,  angle  M^OPi  =  angle  0PM,  and 
OPi  =  OP.     Hence  these  two  triangles  are  equal,  and 

OMj^  =  MP,  M,P,  =  CM. 

The  figures  also  show  that,  for  A  in  each  quadrant,  OJ/j,  MP 
have  the  same  algebraic  sign,  and  JfiA,  OM  have  the  same  sign. 
Hence,  for  all  angles  A, 

M,P,      OM 
OP 


sin  (90°  -  A) 


1^  1  _ 


OM^      MP 


cos  A 


cos  (90  -^)  =  ^  =  _  =  sm  A; 


46.]  RATIOS   OF  180" -A.  81 


tan  (90°  - 

-'')=^=^'="^^ 

cot  (90°  - 

-->'^r"r-'^' 

sec  (90°  - 

-^'-Ji-ip-'""^^ 

osec  (90°  • 

Hence,  the  ratio  of  any  angle  is  the  same  as  the  co-ratio  of  its 
complement.  Compare  with  Art.  16.  The  relations  for  tangent, 
secant,  cotangent,  cosecant,  can  also  be  deduced  from  those  of 
sine  and  cosine  by  means  of  Art.  44  (1),  (2).     Thus,  for  example, 

tan  (90°  -  ^)  =  «-Hli5^^^  =  ^  =  cot  A 
^  ^      cos  (90°-^)      sin^ 

B.  Ratios  of  180°  -  A.  Describe  the  angles  A,  180°  —  A.  Let 
OP,  OPi  be  the  terminal  lines  of  A  and  180°  —  A  respectively. 
In  Figs.  42  a,  42  b,  42  c,  42  d,  A  is  an  angle  in  the  first,  second, 
third,  fourth  quadrants,  respectively.  Take  OPi  equal  to  OP,  and 
draAV  MP,  M^P^,  at  right  angles  to  the  initial  line.  In  the  two 
triangles  OM^P^,  OMP,  in  each  figure,  the  angles  at  M^,  M,  are 


Fia.  42a.  Fia.  42b. 


right  angles,  angle  M^OPx  =  angle  MOP,  OPi=  OP.  Hence, 
0M^=  OM,  and  M^P^=  MP.  In  each  figure,  OM^,  OM  have  oppo- 
site algebraic  signs,  and  M^Pi,  MP,  have  the  same  sign.  Hence, 
for  all  angles  A, 

sin  (180»  -A)  =  ^^=^^AuA; 

/^OAO  ^\  0M^  —    OM  M 

cos  (180°  -  ^)  =  ^  =  —^^  =  -  cos  A. 


82 


PLANE   TRIGONOMETRY. 


[Ch.  V. 


So  also,  tan  (180°  -  ^)  =  -  tan  ^ ;  cot  (180°  -  A)  =  -  cot  ^ ; 
sec  (180°  —  A)  =  —  sec  J. ;  cosec  (180°  —  A)  =  cosec  A. 

The  last  four  relations  can  be  deduced  by  means  of  the  figures, 
or  by  means  of  relations  (1),  (2),  Art.  44.  Hence,  any  ratio  of 
an  angle  is  equal  in  magnitude  to  the  same  ratio  of  its  supplement; 
the  sines  of  supplementary  angles  have  the  same  algebraic  sign,  and 
so  have  the  cosecants;  the  other  ratios  of  supjjlementary  angles  have 
opposite  signs. 

C.  Katios  of  90°  +  A.  Describe  the  angles  A,  90°  +  A.  Let 
OPy  OPi,  be  the  terminal  lines  of  A,  90°  +  A,  respectively.  In 
Figs.  43  a,  b,  c,  d,  A  is  an  angle  in  the  first,  second,  third,  fourth 
quadrants,  respectively.  Take  OPi  =  OF,  and  draw  MP,  M^P^, 
at  right  angles  to  the  initial  line.  In  the  two  triangles,  OMiP^, 
OMP,  (in  each  figure)  the  angles  at  Mu  M,  are  right  angles, 


Fia.  43a. 


Fig.  43b. 


Fia.  43c. 


angle  JfiOPi  =  angle  0PM,  OPi=OP.  Hence,  M^P^=  OM, 
and  OM^  =  MP.  In  each  figure  M^P^,  OM,  have  Ihe  same  alge- 
braic signs,  and  OM^,  MP,  have  opposite  signs.  Hence,  for  all 
angles  A, 

MyPi  ^   OM 

OP 


sin  (90°  +  A) 


OP 


cos    A'y 


COS  (90°  +  ^)  =  I*: 


MP 


=  —  sin  A. 


OP 

So  also,  tan  (90°  +  ^)  =  -  cot  ^ ;         cot  (90°  -h  ^)  =  -  tan  ^ ; 
sec  (90°  4-  ^)  =  -  cosec  A ;  cosec  (90°  +  A)  =  sec  A. 

These  four  relations  can  be  deduced  from  the  figures,  or  by 
means  of  (1),  (2),  Art.  44. 

D.   Ratios  of  -  A.     Describe    the  angles  A,  —  A.     Let   OP, 
OPif  be  the  terminal  lines  of  A,  —  A,  respectively.    In  Figs.  44  a, 


45.] 


RATIOS   OF   —A, 


83 


b,  c,  d,  A  is  in  jhe  first,  second,  third,  fourth  quadrants,  respec- 
tively. Take  OPi  =  OP,  and  draw  PM,  PiM^,  at  right  angles 
to  the  initial  line.  In  the  two  triangles,  OM^P^,  OMP,  (in  each 
figure)  the  angles  at  Mi,  M,  are  right,  angle  M^OPi  =  angle  MOP, 


Fig.  44a. 


Fia.  44b, 


Fig.  44c. 


OPi=OP.  Hence,  MiPi  =  MP,  OM=OMi.  In  each  figure, 
03fi,  OM,  have  the  same  sign,  and  MiP^,  MP,  have  opposite 
signs.     Hence,  for  all  angles  A, 


sin  (—A) 


M,P, 
OP, 


MP 
OP 


=  —  sin  A ; 


,       ..      OMi     OM  . 

eos(-A)  =  ^  =  —  =  cosA. 

So  also,  tan  (—  A)  =  —  tan  A;      cot  (—  ^)  =  —  cot  ^ ; 

sec  (—  ^)  =  sec  ^ ;      cosec  (—  A)  =  —  cosec  A. 

The  last  four  relations  can  be  deduced  from  the  figures,  or  by 
means  of  (1),  (2),  Art.  44. 

Ex.  1.   Show  that  sin  (180°  +  A)=-  sin  A, 

cos  (180°  +  ^)  =  -  cos  ^,  tan  (180°  +  J.)  =  tan^,  etc., 
when  A  denotes  an  angle  in  any  one  of  the  four  quadrants. 

Ex.  2.  Deduce  the  relations  between  -each  of  the  following 
angles  and  angle  A,  viz.  270°  -  A,  270°  +  A,  360°  -  A,  360°  +  A, 
n  •  360°  ±  A,  n  being  any  whole  number. 

By  means  of  the  relations  shoivn  in  this  article,  the  ratios  of  any 
angle  can  be  expressed  in  terms  of  the  ratios  of  an  angle  between 
0°  and  45°.     Thus,  for  example. 


84  PLANE   TRIGONOMETRY.  [Ch.  V. 

sin  700°  =  sill  (360°  +  340°)  =  sin  340°  =  sin  (-  20°)  =  -  sin  20° ; 

tan  975°  =  tan  (2  •  360°  +  255°)  =  tan  255°  =  tan  (180°  +  75°) 

=  tan  75°  =  cot  15° ; 

cosec  (- 1160°)  =  -  cosec  1160°  =  -  cosec  (3  •  360°  +  80°) 

=  -  cosec  80°  =  -  sec  10°. 

.-.  sin  700°  =  -  .34202 ;  tan  975°  =  3.7321 ; 

cosec  (1160°)  =  -  sec  10°  =  -^=J--  =  -=J-  =  - 1.015. 
^  ^  cos  10°     .98481 

This  property,  and  the  property  that  the  ratio  of  an  angle  is 
the  co-ratio  of  its  complement,  account  for  the  arrangement  and 
extent  of  the  trigonometric  tables. 


EXAMPLES. 

1.  Express  the  ratios  of  the  angles  in  Exs.  1-7,  Art.  41,  in  terms  of  ratios 
of  angles  between  0°  and  45°.     Also  find  the  ratios. 

2.  Do  likewise  for  the  angles  in  Exs.  1,  3,  Art.  38.    Also  find  the  ratios. 

3.  Do  likewise  for  the  angles  in  Exs,  1,  2,  3,  Art.  37.    Also  find  the  ratios. 

N.B.    Questions  and  exercises  suitable  for  practice  and  review  on  the 
subject-matter  of  Chap.  V.  will  be  found  at  page  186. 


CHAPTER  VI. 

TRIGONOMETRIC    RATIOS    OF    THE    SUM    AND 
DIFFERENCE    OF    TWO    ANGLES. 

N.B.  Another  way  of  making  the  derivations  shown  in  Arts.  46-48  is 
given  in  Note  B  of  the  Appendix.  The  method  of  projection,  as  it  is  called, 
used  in  Note  B,  is  preferred  by  many. 

46.  Derivation  of  the  sine  and  cosine  of  the  sum  of  two  angles 
when  each  of  the  angles  is  less  than  a  right  angle.  In  this  article 
and  the  following  one,  careful  regard  must  be  paid  to  the  dii-eo- 
tions  in  which  lines  and  angles  are  measured,  and  to  the  order  of 
the  letters  used  in  measuring  them.     See  Arts.  36,  37,  40. 

To  deduce  sin  (A  +  B)  and  cos  (A  +  B).  Let  A  and  B  be  twA 
angles  each  of  which  is  less  than  a  right  angle.  Let  the  turning 
line  revolve  from  the  initial  line  OX,  and  about  0  describe  the 


pI 

/ 

/L 

n-h 

i- 

...-V 

Im\\ 

0      M  N 

X 

Fig.  45a. 

M       O     N 

Fig.  45b. 


angle  XOL  equal  to  A,  and  then  revolve  forward  from  the  posi- 
tion OL  and  describe  the  angle  LOT  equal  to  B.  Thus,  angle 
XOT=A  +  B.  [In  Fig.  45  a,  ^  +  J5  is  less  than  90° ;  in  Fig.  45  h, 
A-\-  B  \^  greater  than  90°.]  Take  any  point  P  on  OT,  the  ter- 
minal line  of  the  angle  {A-{-  B),  and  draw  PQ  at  right  angles  to 
OL,  the  terminal  line  of  the  angle  A.  From  P,  Q,  draw  PM,  QN, 
at  right  angles  to  the  initial  line;  and  through  Q  draw  VQR 
parallel  to  OX  and  intersecting  MP  in  U. 

85 


86  PLANE  TRIGONOMETBY.  [Ch.  VI. 

fA   .    T^        ■     vni:.      ^P      NQ  +  BP      NQ-     RP 

^OQsm^^QPsinFQP      (Definitions,  Art.  40.) 

Now,  by  definitions  in  Art.  40,  and  by  Art.  45, 

^  =  cos  QOP=  cos  i5;  S?=  sin  QOP=  sin  J5; 

sin  VQP  =  sin  (180°  -  PQR)  =  sin  PQR 
=  cos  RQO  =  cos  XOQ  =  cos  A, 
/.  sin  iA  +  B)=  sin  AcosB  -^  cos  A  sin  B,  (1) 

cos(^  +  P)=cosXOP  =  ^=^^^ 


^0^t9)QE^0N     QR 
OP,         OP     OP 

^OQcos^     QPco^VQP 
OP  OP       ' 

(Definitions,  Art.  40.) 

Now,  by  definitions  in  Art.  40,  and  by  Art.  45, 

0^  =  ,,,B,  S^=sinJB; 
OP  'OP  ' 

cosFQP=cos  (180°-PQP)=  -cosPQP=  -sinPQO=  -sin  A 

.'.  cos  (^  +  B)  =  cos  ^  cos  J8  -  sin  A  sin  J5.  (2) 

EXAMPLES. 

1.  Sin  75°  =  sin  (30°  +  45°)  =  sin  30°  cos  45°  +  cos  30°  sin  45° 

=  1      1     ■  \/3      1    ^1+V3 

2  '  V2        2    *  V2        2V2   * 

2.  Find  cos  75°  by  putting  30°  +  45°  for  75°  and  using  formula  (2). 

3.  Deduce  the  sine  and  cosine  of  15°  from  the  results  in  Exs.  1,  2. 


47.]  SIN  (A-B)   AND   COS   {A-B).  87 

-    4.   Find  sin  90°,  cos  90°,  by  putting  90°  =  30°  +  60°.      Also  by  putting 
90°  =  45°  +  45°.     Also  by  putting  90°  =  75°  +  15°. 

5.  Find  sin  120°,  cos  120°,  by  putting  120°  =  60°  +  60°  ;  120°  =  90°  +  30°  ; 
120°  =  75°  +  45°. 

6.  Find  sin  150°,  cos  150°,  by  putting  150°  =  75°  +  75° ;  150°  =  90°  +  60°. 

7.  Find  sin  135°,  cos  135°,  by  putting  135°  =  75°  +  60° ;  135°  =  90°  +  45°. 

8.  Given  sin  x  =  f ,  sin  ?/  =  |,  x  and  y  both  in  the  first  quadrant ;  find 
sin  (x  +  y),  cos(x  +  y). 

47.  Derivation  of  the  sine  and  cosine  of  the  difference  of  two  angles 
when  each  of  the  angles  is  less  than  a  right  angle.  The  construction 
and  derivation  are  veiy  similar  to  that  made  in  the  preceding 
article. 

To  deduce  Hln(A-B)  and  cos(^-jB).  Let  A  and  B  be  two 
angles  each  of  which  is  less  than  a  right  angle,  and  let  A  be  the 
greater.     Let  the  turning  line  revolve  from  the  initial  line  OX, 


^ 

jL 

/T 

^< 

%^/ 

1 

/\^/ 

fp-        y 

k 

t. 

7\ 

0 

N     M 

FIG.  a 

J. 

and  about  0  describe  the  angle  XOX  equal  to  A,  and  then  revolve 
backward  from  the  position  OL  and  describe  the  angle  LOT  equal 
to  -  B.  Then  angle  XOT=A-B.  Take  any  point  P  on  OT, 
the  terminal  line  of  the  angle  (A  —  B),  and  draw  FQ  at  right 
angles  to  OL,  the  terminal  line  of  the  angle  A.  From  P,  Q,  draw 
PM,  QN,  at  right  angles  to  the  initial  line ;  and  through  P  draw 
liPV  parallel  to  OX  and  intersecting  NQ  in  R. 

^  OQ  sin  A      PQ  sin  VPQ 

OP  OP         '      (Definitions;  Art.  40.) 


88  PLANE  TRIGONOMETRY.  [Ch.  VL 

Now,  by  definitions  in  Art.  40,  and  by  Art.  45, 

-^  =  cos  QOP  =  cos  (-  B)  =  cos  B; 

_PQ^QP^^     qOP=  sin(-B)=  -  sin  B: 
OF      OP  ^  ^        ^  / 

sin  VPQ  =  sin  (180°  -  QPR)  =  sin  QPE  =  cos  BQP  =  cos  A. 

.'.  sin  ( 4  -  J5)  =  sin  ^  cos  ^  -  cos  A  sin  B.  (3) 

cos(^-B)  =  cosXOP  =  §f=Mi5^=g|_g 

^OQcos^      PQ  cos  VPQ 

Qp  OP        '    (Definitions,  Art.  40.) 

Now,  by  definitions  in  Art.  40,  and  by  Art.  45, 

— ^  =  cos  B,  and  —  — ^  =  —  sin  Bj  as  shown  above ; 

cosFPQ=cos(180°- QPP)= -cos  QPE=-sinBQP=  -sin A. 
.'.  cos  (A-  B)  =  cos  ^  cos  J5  +  sin  A  sin  B,  (4) 

If  B  is  greater  than  A,  then  the  formula, 

sin  (B  —  A)  =  sin  B  cos  A  —  cos  jB  sin  A, 
can  be  deduced  as  above.    Since 

sin(^  -B)  =  -  sin(B  -  A), 
then  sin  ( J.  —  5)  =  sin  A  cos  5  —  cos  A  sin  ^. 

7i^  is  shoivn  in  Art.  48  that  the  formulas  (1),  (2),  (3),  (4),  are  true 
for  all  values  of  A  and  B.  These  formulas  are  called  the  addition 
and  suhtrojction  formulas  or  theorems  in  trigonometry.  They  are 
of  such  great  importance,  and  so  many  thorems  can  be  deduced 
by  means  of  them,  that  they  are  called  the  fundamental  for- 
mulas of  trigonometry.*     They  should  be  memorized. 

Note.  Arts.  48,  49,  may  be  omitted,  if  deemed  advisable,  until  after  the 
solution  of  triangles  is  completed.    Art.  48  can  also  be  shown  geometrically. 

*  Adrian  Romanus  (1561-1625),  professor  of  mathematics  and  medicine 
at  the  University  of  Louvain,  was  the  first  to  prove  the  formula  for 
sin(^  +  B).  The  formulas  for  cos(^  ±  B)  and  sin(^  —  7?)  were  given  by 
Pitiscus  (1561-1613),  a  German  niatliematician  and  astronomer,  in  his  Trig- 
onometry published  in  1595. 


48.]  ADDITION  AND  SUBTRACTION  FORMULAS.  89 

EXAMPLES. 

1.  Derive  sin  15°,  cos  15°,  on  putting  60°  -  45°  for  15°. 

2.  Derive  sin  15°,  cos  15°,  on  putting  45°  -  30°  for  15°. 

3.  Find  sin  (x  —  ?/),  cos  (x  —  y)  in  tlie  cases  in  Ex.  8,  Art.  46. 

48.  Proof  of  addition  and  subtraction  formulas  for  all  values  of  A 
and  B.  These  formulas  have  been  proved  in  Arts.  46,  47,  for 
values  of  A  and  B  which  are  less  than  a  right  angle.  In  Art. 
45  c  it  has  been  shown  that  for  any  angle,  say  X, 

cos  X  =  sin  (90°  +  X),   sin  X  =  -  cos  (90°  +  X). 

Hence,     cos  ^  =  sin  (90°  +  A),  sin  ^  =  -  cos  (90  °+  A), 

cos(^+i3)  =  sin(90°+^+jB),   sin(.4+^)=-cos(90°+^4--S). 

The  substitution  of  these  values  for  cos  Aj  sin  A,  cos  {A  +  B), 
sin  {A  +  B),  in  (1),  Art.  46,  gives 

-  cos (90°+^+ J5)  =  -cos  (90°+ J.)  cos  5+sin (90°+u4)  sin  5; 


.♦.  cos(90°+^+^)=cos(90°+^)cos5-sin(90°+^)sinJ3.    (1) 
The  substitution  of  the  same  values  in  (2),  Art.  46,  gives 


sin  (90°  +  ^  +  ^)  =  sin  (90°+  A)  cos  B  +  cos  (90°  +  A)  sin  B.    (2) 

Hence,  formulas  (1),  (2),  Art.  46,  are  true  when  one  of  the 
angles  is  increased  by  a  right  angle.  In  a  similar  way,  these 
formulas  can  be  shown  to  remain  true  when  one  of  the  angles  in 
(1),  (2),  of  this  article  is  increased  by  a  right  angle.  It  is  thus 
evident  that  the  formulas  are  true,  no  matter  how  many  right 
angles  are  added  to  either  one  or  both  of  the  angles.  It  can 
easily  be  shown  that  sin  J. = cos  (J.— 90°),  cos  J.  =  — sin  (^—90°). 
Then,  in  the  same  way  as  that  just  emplo'yed,  it  can  be  shown 
that  the  formulas  (1),  (2),  Art.  46,  hold  when  eitlier  one  or  both 
of  the  angles  is  diminished  by  integral  multiples  of  90°.  Hence, 
formulas  (1),  (2),  Art.  46,  are  true  for  angles  in  any  quadrant, 
that  is,  for  all  angles.  In  a  similar  Avay,  formulas  (3),  (4), 
Art.  47,  can  be  shown  to  be  universally  true. 


90  PLANE  TUIGONOMETRY,  [Ch.  VI. 

49.  Each  fundamental  formula  contains  the  others.  From  any 
one  of  the  four  fuildamental  formulas,  the  remaining  three  can 
be  derived.     Thus  for  example : 

In  (1)  Art.  46,  change  A  into  90° -^;  then 

sin  (90°  -  ^  +  5)  =  sin  (90°  -  A)  cos  B  +  cos  (90°  -  A)  sin  B, 
From  this, 


sin  (90°- J.  -  B)f  i.e.  cos  {A-B)=  cos  ^  cos  i?  +  sin  A  sin  B. 

In  (1)  Art.  46,  change  B  into  (— -S);  then 

sin {A  —  B)=  sin ^  cos  (—  5)  +  cos ^  sin (—  B) 
=  sin  A  cos  B  —  cos  A  sin  B. 

In  (1)  Art.  46,  change  A  into  (90°  +  A) ;  then 

sin  (90°  +  -4  +  5)  =  sin  (90°  +  A)  cos  J5  +  cos  (90°  +  A)  sin  B, 

whence,  cos  {A  +  B)=  cos  ^  cos  5  —  sin  A  sin  B. 

Ex.  1.   From  formula  (2),  Art.  46,  derive  the  other  three  fundamental 
formulas. 

Ex.  2.  So  also,  from  formula  (3),  Art.  47. 

Ex.  3.    So  also,  from  formula  (4),  Art.  47. 

50.   Ratio  of  an  angle  in  terms  of  the  ratios  of  its  half  angle. 

In  this  article  and  Arts.  51,  52,  a  few  deductions  will  be  made 
from  the  addition  and  subtraction  formulas,  which  have  been 
shown  to  be  true  for  all  angles.  These  deductions  are  necessary 
for  the  explanations  concerning  triangles,  as  well  as  useful  for 
other  purposes.  More  ample  opportunity  will  be  afforded  later 
for  working  exercises  involving  the  use  of  these  formulas.  The 
fundamental  formulas  may  be  brought  together: 

sin  {A-\-  B)-  sin  A  cos  B  +  cos  vl  sin  B.  (1) 

sin  {A-  B)  =  sin  ^  cos  B  -  cos  A  sin  -B.  (2) 

cos(^  +  B)  =  cos  A  cos  B  -  sin  A  sin  B.  (3) 

cos(^  -  B)  =  cos  ^  cos  B  +  sin  A  sin  J5.  (4) 


49-50.]  SIN  2  A,    COS  2  A.  91 

Let  B  =  A;  tlien,  from  (1), 

sin  (A  +  A)=  sin  A  cos  A -\-  cos  A  sin  A  j 

that  is,  sin  2  ^  =  2  sin  A  cos  A.  (5) 

Similarly,  from  (3),    cos  2  ^  =  cos*^  A  -  sin^^.     n  (6) 

Since  cos^  A  +  sin^  J.  =  1,  it  follows  that 

cos2^  =  l-2sin2^.  (7) 

and  cos  2 4  =  2  cos^  A-  1.  (8) 

In  formulas  (l)-(8),  A,  B,  denote  any  angles  whatsoever. 
These  formulas  occur  so  often,  and  are  so  useful,  that  it  is  well 
to  translate  them  into  words.     Thus, 

sine  sum  of  any  two  angles  =  sin  j^rs^  •  cosine  second 

+  cosine  first  •  sine  second 
sine  difference  of  any  two  angles  =  sine  j^Vs^  •  cosine  secoyid 

—  cosine  first  •  sine  second 
cosine  sum  of  any  two  angles  =  cosine  first  •  cosine  second 

—  sine  j^Vs^  •  sine  second 
cosine  difference  of  any  two  angles  =  cosine  first  •  cosine  second 

+  sine  first  •  sine  second 

Since  A  is  one-half  of  2^4,  formulas  (5)-(8)  can  be  translated 
as  follows : 

sine  any  angle  =  2  sine  half-angle  •  cosine  half-angle, 

cosine  any  angle  =  (cosine  half-anglef  —  (sine  half-angle)'"^, 
=  1  —  2  (sine  half-angley, 
=  2  (cosine  half-angle^  —  1. 

EXAMPLES. 

1.   Find  cos  221°  from  cos  45°. 

2  cos2  22|°  =  1  +  cos  45°  by  (8)  ; 

^  V         V2/         2V2        2x  1.4142 

.-.  cos22i°  =  .9239. 


92  PLANE  TRIGONOMETRY.  [Ch.  VI. 

2.  Express  cos  4  x  in  terms  of  sin  x  and  cos  x. 

cos4ic  =  2cos2  2x  -  1  =  2(2cos2x-  1)2-  1  =  8  cos'* x  -  8  cos2 cc  +  1. 

3.  Deduce  sin  30°,  cos  30°,  from  cos  60°. 

4.  Deduce  sin  75°,  cos  75°,  from  cos  150°.     [Logarithms  may  be  helpful.] 

5.  Deduce  sin  22|°,  from  cos  45°. 

6.  Deduce  sin  15°,  cos  15°,  from  cos  30°. 

7.  Express  cos  6  x,  sin  6  x,  in  terms  of  ratios  of  3  x. 

8.  Express  cos  3  x,  sin  3  x,  in  terms  of  ratios  of  f  x. 

9.  Express  sin  f  a*,  cos  |  x,  in  terms  of  ratios  of  f  x. 

10.  Express  cos  6  x,  sin  6  x,  in  terms  of  ratios  of  12  x. 

11.  Express  cos  3  x,  sin  3  x,  in  terms  of  ratios  of  6  x. 

12.  Express  sin  |  x,  cos  |  x,  in  terms  of  ratios  of  |  x. 

13.  Show  that  sin  (w  +  1)^  +  sin  (w  -  1)  J.  =  2  sin  nA  cos  A,  and 

cos  (n  +  V)A  +  cos  {n  —  1)^  =  2  cos  nA  cos  A  ;  and 
hence,  express  sin  2  ^,  cos  2  ^,  in  terms  of  sin  A^  cos  ^. 

51.  Tangents  of  the  sum,  and  difference  of  two  angles,  and  of 
twice  an  angle.  Let  Ay  B,  be  any  two  angles.  It  is  required 
to  find  tan  {A  +  B)  and  tan  {A  —  B). 

tan  (A-\-B)  =  ^^^  (^  +  B)  _  sin  ^  cos  -S  +  cos  A  sin  B 
cos  {A  +  B)      cos  ^  cos  ^  —  sin  ^  sin  B 

On  dividing  each  term  of  the  numerator  and  the  denominator 
of  the  second  member  by  cos  A  cos  B,  there  is  obtained 

tan(^  +  B)=:  tan^  +  tang,  (1) 

1  -  tan  ^  tan  ^ 

In  the  same  way  it  can  be  shown  that 

tan(^-B)^  *^»/-;f  ^.  (2) 

1  +  tan  ^  tan  B 

Formula  (2)  can  also  be  deduced  from  (1)  by  changing  B  into 
-B. 
If  B  =  A,  then  (1)  becomes 

tan2^  =  ^iM.^.  ^        (3) 

1  -  tan^  A 


51-52.3  SUM  OF  TWO   SINES.  93 

Formulas  (1),  (2),  (3),  can  be  translated  into  words,  as  follows : 

tangent  sum  any  two  angles  = ^ ^ ; 

1  —  product  of  tangents 

tangent  difference  any  two  angles  =   difference  of  tangents 

1  +  product  of  tangents 

tangent^angle^^fg;^;;;^^, 

EXAMPLES. 

1.  tan  P  =  2,  tan  §  =  f     Find  tan  (P+  Q),  tan  (P  -  Q). 

t 
tan(P+§)=:-l±l^  =  7;    tan  (P -  ^)  =  Azii^  =  i. 

2.  Find  tan  75°  by  means  of  tan  4^°,  tan  30°. 

3.  Find  tan  15°  by  means  of  tan  60°,  tan  45°. 

4.  Find  tan  22|°  from  tan  45°.  5.   Find  tan  37|°  from  tan  75°. 

6.  Derive  cot  iA±B)  =  ^ot^cotP:Fl    .^^  2  ^  =  co^^jzi. 
cot  J5  ±  cot  ^  2  cot  A 

52.   Sums   and   differences   of   sines  and   cosines.      The   set   of 

formulas  (l)-(4),  Art.  50,  can  be  transformed  into  two  other  sets 
which  are  very  useful.  From  (1),  (2),  (3),  (4),  Art.  50,  on  addi- 
tion and  subtraction,  there  is  obtained : 

sin(^  +  B)+sin(^-.B)=2sin^cosJ5.  (1) 

sin  (A  +  B)-  sin  (A-B)=2  cos  A  sin  B.  (2) 

cos  (A  +  B)-{-  cos  iA-B)=2  cos  A  cos  B.  (3) 

cos  iA  +  B)-  cos  (^  -  ^)  =  -  2  sin  ^  sin  B,  (4) 

If  A  +  B  =  P, 

and  A  —  B=  Q, 

then  2  ^  =  P+  Q,  and  ^  =  i  (P^+  Q), 

2B==P-  Q,  and  JS  =  i(P-  Q). 


9^  PLANE  TRIGONOMETBY.  [Ch.  VI. 

Substitution  of  these  values  of  A,  B,  in  (l)-(4)  gives 

siu  r+s'mQ=     2  sin  ^-^  cos  ^^-  (5) 

sin  P-  sin  <?  =     2  cos  ^±-^  sin  ^^^  (6) 

cos  P  +  COS  (?  =     2  cos  ^^  ^  cos  ^^=-^«  Ci") 

COS  P-  COS  Q  =  -  2  sin  ^^  sin  ^^-  (8) 

Formulas  (l)-(4)  with  the  members  transposed,  are  useful  for 
transforming  products  of  sines  and  cosines  into  sums  and  differ- 
ences; formulas  (5)  to  (8)  are  useful  for  transforming  sums  and 
differences  of  sines  and  cosines  into  products.  These  formulas 
may  be  translated  into  words :  —  Of  any  two  atigles, 

2  sin  one  •  cos  the  other  =  sin  sum  +  sin  difEerence,  .  .  .  (!') 

2  cos  one  •  sin  the  other  =  sin  sum  —  sin  difference,  .  .  .  (2') 

2  cos  one  •  cos  the  other  =  cos  sum  +  cos  difference,  .  .  .  (3') 

2  sin  one  •  sin  the  other  =  cos  difference  —  cos  sum,  .  .  .  (4') 

the  sum  of  two  sines  =  2  sin  half  sum  •  cos  half  difference,     (5')  " 
the  difference  of  two  sines  =  2  cos  half  sum  •  sin  half  difference,     (6') 
the  sum  of  two  cosines  =  2  cos  half  sum  •  cos  half  difference,     (7') 
the  difference  of  two  cosines  =  —  2  sin  half  sum  •  sin  half  difference.  (8')  . 

The  difference  between  the  first  members  of  A,  Art.  50,  and 
C  should  be  noted. 

N.B.  Arts.  92-95  are  similar  in  character  to,  and  are  merely  a  continua- 
tion of,  Arts.  50-52.  If  deemed  advisable,  Arts.  91-95  can  be  taken  up 
now.  The  student  is  advised  to  glance  at  them  after  solving  the  following 
exercises : 

EXAMPLES. 

1.   tfhowthat  ^28^^1^251  =  _  tan  |(a; +  2/)  tan  1  (a;- y). 
cos  x  + cosy 

cosx- cosy  ^-2sinKa^  +  y)sinKa;-.v)^_tani(x  +  y)  tan  Ka;-y). 
cosx+cosi/       2cos\{x+y)co&\{x-y)  ^K-^-rs/J^    Ik*      !fJ 


52 

.] 

EXAMPLES. 

( 

2. 

Show  that      ^^^^^     =  tan  A. 
1  +  cos  2  ^ 

sin  2  ^              2  sin  A  cos 
1  +  cos  2  ^      1  +  (2  cos2  A 

A 

-1) 

sin^ 
cos^ 

tan  A 

3. 

Show  that  2  sin  (A  +  45°)  sin  (^  - 

-45'^): 

=  sin2  A 

-  cos2  ^. 

95 


2  sin  (^ + 45°)  sin  {A  -  45°)  =  cos  (^ + 45°  -  ^  -  45°)  -  cos  (^ + 45° + ^  -  45^) , 

Art.  52,  B  (40 
=  cos  90°  —  cos  2  ^  =  sin2  A  —  cos^  A. 

4.   Show  that  sin  5  A  sin  A  =  sin2  3  ^  —  sin2  2  A. 

sin  5  ^  sin  ^  =  i  [cos  (5  ^  -  ^)  -  cos  (5  ^  +  ^)],  Art.  52,  B  (4') 

=  ^  (cos  4  J.  —  cos  6^), 
=  Kl-2sin22^-(l-2sin23^)]  =  sin23^-sin22A 

6.   «yLA±^HLM  =  tan2A 
cos  A  +  cos  3  A 

sin  ^  + sin  3^  _  2  sin  \(SA  +  A)  cos ^(SA-A)      Art  52  C  (6')  (T^ 
cos^  +  cos3^     2cosK3^  +  ^)cosK3^-^)'  *      '     "<    ^'"^   ^ 

=  siEli  =  tan2A 
cos  2  A 

Prove  the  following  statements : 

g    sin^+sinB _  tan^(^  +  ^)  -    sin  3a;  — sin  a; _  ^^^ ^ 

sin^— sinJ5     tan^(^— J5)  *   cos3a!  +  cosaj 

8.  cos  (A  +  B)  cos  (A-  B)=  cos2  A  -  sin2  J5. 

9.  cot  ^  —  cot  2  ^  =  cosec  2  A. 

10.  sin  (A  +  B)  sin  (A-  B)  =  cos2  5  -  cos^  A, 

11.  1  +  tan  2^  tan  ^  =  sec  2  A 

12.  tan(45°  +  ^)  =  l-±^5:5^.  13.   tan  (45°  -  ^)  =  ^  ~  ^^^  ^- 

^  ^     1-tan^  ^  ^     1  +  tan^ 

14.    fsin  — +  C0S— V=  1  +  sinA       15.    fsin  —  -  cos  —  )  =  1  -  sin  A 
V       2  2;  ^  V       2  2; 

16.   hl-^2^1A  =  t^nA.  17.  -^HiM_  =  cotA 


sin  2  J.  1  —  cos  2  ^ 

±^2^  =  cot^.  19.   -ll^IL^ 

sin  ^  2  1  +  tan2  A 


96  PLANE   TIUGONOMETRY.  [Cu.  Vl. 

_       cosec^         ^  gee 2  A  21.  .^  "  ^""'^  =  cos2  A 


cosec2  A  — 2,  sec2  ^ 

22.   sin(^  +  ^)^tan^  +  tan^.        23.   ^HLl^^l^  =  tan  4  -  tan  5. 
cos  ^  COS  B  cos  J.  cos  J5 

24.   22^MzL^  =  cotJ5  +  tanA        25.   £2^1^.+^  =  cot  ^  -  tan  ^. 
cos  ^  sin  B  sin  ^  cos  ^ 

2g    sin  (a;  +  y)  _  tan  x  +  tan  y  07    cos  (x  +  y)  _  1  —  tan  x  tan  y 

sin  (X  —  y)      tan  x  —  tan  y  cos  (x  —  y)      1  +  tan  x  tan  ?/ 

28.  Given  sin  x  =  f ,  sin  y  =  |.  Find  sin  (x  +  ?/),  sin  (x  —  y),  cos  (x  +  y), 
cos(x  — y),  sin2x,  sin  2?/,  cos2x,  cos  2  2/,  tan2x,  tan  2?/,  tan(x  +  y), 
tan  (x  —  y),  when  (a)  both  x,  y,  are  in  the  first  quadrant ;  (6)  x  is  in  the  first, 
y  in  the  second ;  (c)  x  in  the  second,  y  in  the  first ;  (d)  both  in  the  second 
quadrant.    Check  results  by  means  of  the  tables. 

29.  Given  sin  x  =  ^,  sin  y  =  f .     Do  as  in  Ex.  28. 

30.  Given  sin  x  =  f ,  sin  y  =  —  f .  Find  the  ratios  named  in  Ex.  28,  when 
X  is  in  the  first  quadrant  and  y  in  the  third,  x  in  the  first  and  y  in  the  fourth, 
X  in  the  second  and  y  in  the  third,  x  in  the  second  and  y  in  the  fourth. 

N.B.  Examples  suitable  for  exercise  and  review  on  the  subject-matter  oj 
this  chapter  will  be  found  in  Arts.  91-95,  and  at  page  187. 


CHAPTER   VIL 

SOLUTION   OF  TRIANGLES    IN   GENERAL. 

53.  Cases  for  solution.  In  Art.  34  oblique  triangles  were  solved 
by  means  of  right-angled  triangles.  In  this  chapter  some  rela- 
tions of  the  sides  and  angles  of  any  triangle  (whether  right-angled 
or  oblique)  will  be  derived ;  methods  of  solution  will  be  shown, 
which  are  applicable  to  the  solution  of  both  right-angled  and 
oblique  triangles,  and  which  are  independent  of  the  special  aid 
that  can  be  afforded  by  right-angled  triangles.  In  Art.  54  the 
chief  relations  between  the  sides  and  angles  of  a  triangle  will  be 
deduced.  These  relations  constitute  the  foundation  for  the  re- 
mainder of  the  chapter.  In  Arts.  55-58  solutions  of  triangles 
are  obtained  without  the  use  of  logarithms ;  in  Arts.  60-62  loga- 
rithms are  employed  in  finding  the  solutions. 

In  order  that  a  triangle  may  be  constructed,  three  elements,  one 
of  which  must  be  a  side,  are  required.  Hence,  there  are  four 
cases  for  construction  and  solution,  namely,  when  the  given  parts 
are  as  follows : 

I.  One  side  and  two  angles. 

II.  Two  sides  and  the  angle  opposite  to  one  of  them. 

III.  Two  sides  and  their  included  angle. 

IV.  Three  sides. 

Before  proceeding,  the  student  should  test  his  ability  to  con- 
struct a  triangle  readily  in  each  of  these  cases. 

In  the  discussions  that  follow,  the  triangle  is  denoted  by  ABC, 
the  angles  by  A,  B,  O,  and  the  lengths  of  their  opposite  sides  by 
a,  6,  c,  respectively.* 

*The  formulas  are  greatly  simpljfied  by  the  adoption  of  this  notation, 
which  was  first  introduced  by  Leonhard  Euler  (1707-1783). 

97 


98 


PLANE  TRIGONOMETRY. 


[Ch,  VII. 


54.  Fundamental  relations  between  the  sides  and  angles  of  a 
triangle.     The  law  of  sines.     The  law  of  cosines. 

I.  The  law  of  sines.  From  C  in  the  triangle  ABC  draw  CD  at 
right  angles  to  opposite  side  AB,  and  meeting  AB  or  AB  produced 
in  D.     (In  Fig.  47  a  B  is  acute,  in  Fig.  47  6  £  is  obtuse,  and  in 


A     c    BD~V         A^^^SZB^^ 


AD  B  V       A     ^    B      D  V 

Fig.  47a.  Fig.  47b.  Fig.  47c.  Fig.  48. 


Fig.  47  c  jB  is  a  right  angle.)    Produce  AB  to  V.    In  what  follows, 
AB  is  taken  as  the  positive  direction. 

In  CD  A,  DC  =b  sin  A. 

In  CDB  (Figs.  47  a,  b),   DC  =  a  sin  VBC   (Definition,  Art.  4#.) 

=  a  sin  B. 
['.'  sinF^a=  sin(180°  -  VBC),  Art.  45,  =  sin  CBA] 

In  Fig.  47c,  DG=BC=a  =  a sinB. 

('.'  B  =  90°,  and  sin  90°  =  1) 
Therefore,  in  all  three  triangles, 

a  sin  B=  b  sin  A. 

b 


Hence, 


a ^ 

sin  A     sin  B 


Similarly,  on  drawing  a  line  from  B  at  right  angles  to  AC,  it 
can  be  shown  that 

g     _     c 
sin  A     sin  G 

Hence,  in  any  triangle  ABC, 


sin  A     sin  B     sin  C 


(1) 


54.] 


LAW  OF  SINES. 


99 


In  words :  TJie  sides  of  any  triangle  are  propoHional  to  the  sines 
of  the  opposite  angles. 

Each  of  the  fractions  in  (1)  gives  the  length  of  the  diameter  of 
the  circle  described  about  ABC.  Let  0  (Fig.  48)  be  the  centre  and 
R  the  radius  of  the  circle  described  about  ABC.  Draw  OD  at 
right  angles  to  any  one  of  the  sides,  say  AB.  Draw  AO.  Then 
AD  =  \c,  ADD  =  C,  by  geometry.     In  triangle  AODj 


AD  =  AO^mAOD]  i.e.  ^c  =  Rs>mC. 


.•.2U 


sin  C 


(2) 


Ex.  1.   Explain  why  the  circumscribing  circle  of  a  triangle  depends  only 
'upon  one  side  and  its  opposite  angle. 

Ex.  2.   Derive  the  law  of  sines  by  drawing  a  perpendicular  from  ^  to  J5C. 

Ex.  3.   Derive  2B  =  — ^,  2B  =  — ^,  by  means  of  figures, 
sin  A  sin  B 

II.  The  law  of  cosines.  An  expression  for  the  length  of  the 
^  side  'of  a  triangle  in  terms  of  the  cosine  of  the  opposite  angle 
and  the  lengths  of  the  other  two  sides,  will  now  be  deduced. 
The  angle  A  is  acute  in  Fig.  49  a,  obtuse  in  Fig.  49  b,  right  in 
Fig.  49  c.  From  C  draw  CD  at  right  angles  to  AB.  The  direc- 
tion AB  is  taken  as  positive. 


In  Figs.  49  a,  49  6, 
In  Fig.  49  a, 
in  Fig.  49  b, 


BC'  =  DC'  +  DB", 
DB  =  AB-AD) 
DB=DA-\-AB  =  -AD-\-  AB. 


Hence,  in  both  figures,  BC  =  DC  +  (AB  -  ADy 


DC  +  AD-  +  AB'  -2AB'  AD. 


100  PLANE  TBIGONOMETRY.  [Ch.  VII. 

In  Fig.  49  a,  AD  =  AC  cos  BAC; 

in  Fig.  49  6,  AD  =  ^(7  cos  BAC  (Art.  40). 

Also,  DO'  +  Aff=ACf. 

Hence,  in  both  figures,  W  =  J^O'  +  Zb' -2  AC-  AB cos ^ ; 

tliat  is,  a2  :=  62  4-  c2  -  2  6c  cos  ^.  (3) 

This  formula  also  holds  for  Fig.  49  c ;  for  there, 

cos  ^=:  cos  90°  =  0. 

Similar  formulas  for  6,  c,  can  be  derived  in  like  manner,  or  can 
be  obtained  from  (3)  by  symmetry : 

62  =  c2  +  a2-2cacos^,  c"  =  o? -\-h'' -2ah(tosC.         (3') 

These  formulas  can  be  expressed  in  words :  In  any  triangle,  the 
square  of  any  side  is  equal  to  the  sum  of  the  squares  of  the  other  two 
sides  minus  tivice  the  product  of  these  two  sides  multiplied  by  the 
cosine  of  their  included  angle. 

Note.  In  Fig.  49  at,  A  is  acute  and  cos  A  is  positive  ;  in  Fig.  49  b,  A  is 
obtuse  and  cos  A  is  negative.  Hence  formula  (3)  shows  that  in  Fig.  49  a, 
a^  is  less  than  6^  +  c^,  and  that  in  Fig.  49  b,  a'^  is  greater  than  b^  +  c^.  In 
Fig.  49  c,  a2  =  62  +  c2. 

-<« 

Relation  (3)  may  be  expressed  as  follows : 

oosA  =  ^±^,  (4) 

and  similarly  for  cos  B,  cos  O. 

Ex.   Derive  the  formulas  for  6^  and  for  c*. 

Each  of  the  relations  (1),  (3),  (3'),  involves  four  of  the  six  ele. 
ments  of  a  triangle.  If  any  three  of  the  elements  in  any  one 
of  these  relations  are  known,  then  the  fourth  element  can  be 
found  by  solving  the  equation.  Inspection  shows  that  relations 
(1)  are  serviceable  in  the  solution  of  Cases  I.,  II.,  Art.  53,  and 
that  relations  (3),  (3'),  are  serviceable  in  the  solution  of  Cases  III., 
IV.,  Art.  53.  The  student  is  advised  to  try  to  work  some  of  the 
examples  in  Arts.  55-58  before  reading  the  text  of  the  articles. 
(See  Arts.  20-24,  34.) 


54  a,  55.]  SOLUTION   OF  TRMNGLES:.   CASE  I-  101 

54  a.    Substitution  of  sines  for  sides,  and  of  sides  for  sines. 

Since  — ^  =  — —  =     ^    ,  the  sines  of  the  opposite  angles  can  be  sub- 
sin  A     sin  B     sin  C 

stituted  for  the  sides  of  triangles,  and  vice  versa,  when  they  are  involved 

homogeneously  in  the  numerator  and  denominator  of  a  fraction,  or  in  both 

members  of  an  equation. 

Thus,  on  putting  — ^  =  — ^  =  — ^  =  x,  it  follows  that  a  =  x  sin  A^ 
sin  A     sin  B     sm  G 
b  =  xsmB,  c  =  x sin C. 

a^  ax  sin  A  a  sin  ^ 


Then,  for  example, 


6  +  c     ic  sin  ^  +  ir,  sin  (7     sin  JB  +  sin  G 


EXAMPLES. 


1.  Show  that  in  any  triangle  ^±^  =  ^^^-^), 

c  sm  I  C 

a  +  5_sin^  +  sin  ^  _  2  sin |(yl  +  ^)cosK^  —  -^)  _cos^(^  —  -S) 
c     "~        sin  C  2  sin  I C  cos  ^G  sin  ^  O       ' 

for  sin  l(A  +  B)=  cos  |  O,  since  1(A  +  B)  +  ,IG  =  90°. 

2.  Derive  two  other  relations  similar  to  that  in  Ex.  1. 

3.  Show  that 

3 qg  4.  2  62 _  3sin2^  +  2sin2^  _  3 sin^  A +  2  sin^ B _ 3 sin^ ^4  +  2 sin^^ .g 
abc       ~      a  sin  ^  sin  O  6  sin  ^  sin  O  c  sin  J.  sin  J? 

55.  Case  I.  Given  one  side  and  two  angles.  In  triangle  ABC, 
suppose  that  A,  B,  a  are  known ; ,  it  is  required  to  find  C,  b,  c. 
In  this  case  (see  Fig.  47  a,  Art.  54), 

C=1S0°-(A  +  B)', 

,  whence  b  =   .^  ,  •  sin  5 : 
sm^ 


_  •,  whence  c  =  -: — -  •  sin  O. 

sin  C      sm  A  sm  A 

b  c 

b^  -{-  c^  —  2bc  cos  A,    — =  -: — -,  the   result  in 

sm  B     sm  C 

Ex.  1,  Art.  54  a.     Other  checks  will  be  discovered  later. 


h 

a 

sin  5 

sin  J.' 

c 

a 

102 


:p^ANK  TMIGONOMETBT. 


[Ch.  VII. 


EXAMPLES. 
1.   Solve  the  triangle  PQB,  given  :  PQ  =  12  in., 


Q  =  40°, 

P  =  75°. 

B  =  180°  -  (P  +  ^)  =  180°  -  (40°  +  75°)  =  65°. 

&\nQ     sini?  "  sinP     sini? 


Solution  :  B 
PB 


BQ  = 


sin  JK 

12 
sin  65 

12 


•  sin  Q, 
;.sin40°, 
X  .  6428, 


P^  =  -^.sinP, 
sinP 


12 


sin  65° 
12 


.  sin  75° 


X  9659, 


.9063'             '  .9063 

=  13.24  X  •  6428,  =  13.24  x  9659, 

=  8.51  in.  =  12.8  in. 

2.  In  ABC,  A  =  50°,  B  =  75°,  c  =  60  in.     Solve  the  triangle. 

3.  In  ABC,  A  =  131°  35',  B  =  30°,  b  =  ^  ft.  Find  a. 

4.  In  ^PO,  B  =  70°  30',  C  =  78°  10',  a  =  102.  Solve  the  triangle. 

5.  In  ABCy  B  -  98°  22',  C  =  41°  1',  a  =  5.42.  Solve  the  triangle. 

56.   Case  II.     Given  two  sides  and  an  angle  opposite  to  one  of 

them.     In  the  triangle  ABC  let  a,  h,  A  be  known,  and  G,  Bj  c  be 
required.     The  triangle  will  first  be  constructed  with  the  given 


b 

G 

/ 

A 

/ 

a 

/ 

/       c. 



Fig.  50. 


Fig.  51. 


b                   ^C 

\/      /      I     c         \ 

s/- 

P;\  A        D 
FlO.  50. 


^^'B  M 


56.]  SOLUTION    OF  TRIANGLES :    CASE  It.  103 

elements.  At  any  point  J.  of  a  straight  line  LM,  unlimited  in 
length,  make  angle  MAG  equal  to  angle  A,  and  cut  off  AG  equal 
to  h.  About  (7  as  a  centre,  and  with  a  radius  equal  to  a,  describe 
a  circle.     This  circle  will  either : 

(1)  Not  reach  to  LM,  as  in  Fig.  50. 

(2)  Just  reach  to  LM,  thus  having  LM  for  a  tangent,  as  iu 
Fig.  51. 

(3)  Intersect  LM  in  two  points,  as  in  Figs.  52,  53. 

Each  of  these  possible  cases  must  be  considered.  In  each 
figure,  from  G  draw  GD  at  right  angles  to  AM]  then  GD  =  b  sin  A. 

In  case  (1),  Fig.  50,  GB  <  GD,  and  there  is  no  triangle  which 
can  have  the  given  elements.  Hence,  the  triangle  is  impossible 
when  a  <b  sin  A. 

In  case  (2),  Fig.  51,  GB  =  GD.  Hence,  the  triangle  which  has 
elements  equal  to  the  given  elements  is  right-angled  when  a  =  b  sin  A. 

In  case  (3),  Figs.  52,  53,  GB>GD',  that  is,  a>b  sin  A.  If 
a  >  b,  then  the  points  B,  B^,  in  which  the  circle  intersects  LM, 
are  on  opposite  sides  of  A,  as  in  Fig.  52,  and  there  is  one  triangle 
tvhich  has  three  elements  equal  to  the  given  elements,  namely,  ABG. 
Jf  a<b,  then  the  points  of  intersection  B,  B^,  are  on  the  same 
side  of  A,  as  in  Fig.  53,  and  there  are  two  triangles  which  have 
elements  equal  to  the  given  elements,  namely,  ABG,  AB^G.  For,  in 
ABG,  angle  BAG=A,  AG=  b,  BG==a;  in  AB,G,  angle  B^AG=A, 
AG  =  b,  BiG  =  a.  Both  triangles  must  be  solved.  In  this  case. 
Fig.  53,  the  given  angle  is  opposite  to  the  smaller  of  the  two 
given  sides.  Hence,  there  may  be  two  solatio7is  wheri  the  given 
angle  is  opposite  to  the  smaller  of  the  two  given  sides.  The  words 
"  7nay  fte"  are  used,  for  in  cases  (1),  (2),  the  given  angle  is 
opposite  to  the  smaller  of  the  two  given  sides.  Case  II.  is  some- 
times called  the  ambiguous  case  in  the  solution  of  triangles. 

The  ambiguity  in  Gase  II.  is  also  apparent  in  the  trigonometric 
solution.     The  angle  B  is  found  by  means  of  the  relation, 

sin  J5     sin  ^  -     -n      I)   •     a  m\ 

= :  or,  sm  5  =  -  sm  A.  (1) 

b  a  a  ^ 

The  angle  B  is  thus  determined  from  its  sine.  Now  there  is 
always  an  ambiguity  when  an  angle  of  a  triangle  is  determined 


104 


PLANE  TRIG ONOMETR  Y. 


LCh.  VII. 


from  its  sine  alone,  for  sin  x  =  sin  (180°  —  x).  Figure  53  shows 
the  two  angles  which  have  the  same  sine,  namely,  ABC,  AB^C. 
In  Fig.  52,  the  given  condition,  namely,  that  b  <a,  shows  that 
B<A]  accordingly,  only  the  acute  angle  corresponding  to  sin  B 
can  be  taken.  If,  in  equation  (1),  bsin  A>  a,  then  sin  B>1, 
and,  accordingly,  B  is  impossible  and  there  is  no  solution.  If,  in 
equation  (1),  bsin  A  =  a,  then  sin  ^  =  1,  and  B  =  90°.  The  con- 
sideration of  the  trigonometric  equation  (1)  leads,  therefore,  to 
the  same  results  as  the  preceding  geometrical  investigation. 

Checks:  A-{-B-\-  (7=180°,  and,  as  in  Case  I.     Other  checks 
will  be  found  later. 

EXAMPLES. 
1.    Solve  the  triangle  STV,  given  :  ST  =  15,  VT  =\2,  S=  52°. 

sin  V_smS .  ^T 

VT  ' 

.78801 


8T 

sin  V     sin  52° 


=  .065668. 


15  12  12 

.-.  sin  F=  15  X  .065668  =  .98502. 
.-.   F=  80°  4'  20",  or  180°  -  80°  4'  20",  i.e.  99°  55'  40". 
Both  values  of  V  must  be  taken,  since  the  given  angle 


S  V^        V 

EiG.  54. 

is  opposite  to  the  smaller  of  the  given  sides.     The  two  triangles  correspond- 
ing to  the  two  values  of  Fare  STV,  S2'V\,  Fig.  54,  in  which 

/SFr=  80°  4' 20",    >S'Fir=99°55'40". 


In  STVi 

In  STV 

angle  STV^  =  180°  -  (^  +  SV^T) 

angle  STV=  180°  -(6^+  8VT) 

=  28°  4'  20". 

=  47°  55'  40". 

SVx      _VxT  . 
smSTVi     sin/S' 

SV        VT  . 
sin  STV    smS* 

.47059 -.78801 -^^•^^^• 

'^^  =  15.228. 
.74230 

.-.  SVi  =  lAl. 

.-.  >S'F=11.3. 

The  solutions  are  : 

Fi  =  99°  55'  40"  ■ 

F=80°    4' 20" 

T=28°    4' 20" 

> 

r=  47°  55' 40" 

/S'Fii^7.17 

^F=11.3 

In  the  ambiguous  case,  care  must  be  taken  that  the  calculated  sides  and 
angles  are  combined  properly. 


57.]  SOLUTION  OF  TRIANGLES :    CASES  III.,   IV.  105 

2.  Solve  ABC,  given  :  a  =  29  ft.,  6  =  34  ft.,  ^  =  30°  20'. 

3.  Solve  ABC  when  a  ^  30  ft.,  &  =  24  ft.,  5  =  65°. 

4.  Solve  ABC  when  a  =  30  in.,  6  =z  24  in.,  ^  =  65°. 

5.  Solv^  ABC  when  a  =  15  ft.,  &  =  8  ft.,  J5  =  23°  25'. 

57.  Case  III.  Given  two  sides  and  their  included  angle.  In  the 
triangle  ABC,  a,  b,  C  are  known,  and  it  is  required  to  find  A, 
B,  c.  In  this  case,  c  can  be  determined  from  the 
relation  c^  =  a^  -^b^  —  2ab  cos  (7,  Art.  54 ;  angle 
A  can  be^  determinedrfromihe  relation 

sin  A_suiC . 
a  c 

angle  B  can  be  determined  from  the  relation 

^  +  5+ (7=180°,  or  from  ^=^HL^. 

b  c 

Checks:  ^a^  =  W  -\- c^  —  2  be  cos  A,  b^  ==a^  4-  c-  —  2 ac  cos  A^  the 
result  in  Ex.  1,  Art.  54  a ;  other  checks  will  be  found  later. 

EXAMPLES. 

1.    In  triangle  PQIi,  p  =  8  ft.,  r  =  10  ft.,  Q  =  47°.     Find  q,  P,  B. 
q2  —p2  ^  ^2  _  2pr  cos  Q 

=  64  +  100  -  2  X  8  X  10  X  .6820  =  54.88. 
.-.  q  =  7.408. 

sin  P  =  P^rL9.  =  8j^^7314  ^  ^g^g^      ._  p  ^  ^^o  ^q, 
q  7.408 

sin  B  =  ri^HLS  =  ^»  X  -7314  ^  gg^g       ._  ^  ^  g^o  50/. 
g  7.408 

2.  Solve  ylPC,  given  :  a  =  34  ft.,   &  =  24  ft.,    C  =  59°  17'. 

3.  Solve  ABC,  given  :  a  =  33  ft.,   c  =  30  ft.,  B  =  35°  25'. 

4.  Solve  J^/ST,  given  :    r  =  30  ft.,    s  =  54  ft.,   r=46°. 

6.   Solve  PQB,  given  :  p  =  10  in.,  g  =  16  in.,  B  =  97°  54'. 

58.  Case  IV.  Three  sides  given.  If  the  sides  a,  b,  c  are  known 
in  the  triangle  ABC,  then  the  angles  A,  B,  C  can  be  found  by 
means  of  the  relations  (3),  Art.  54. 

Checks:  Kelations  (1),  Art.  54:  A  +  B-^C^180°.  Other 
c'hecks  will  be  shown  later. 


106  PLANE  TRIGONOMETUr,  [Ch.  VII. 

EXAMPLES. 
1.   In  ABC,  a  =  4,  6  =  7,  c  =  10  ;  find  ^,  B,  C. 

cos  A  =  yL±^J^ot  ^  49  +  100-16  ^  133  ^  ^^q^.    .-.  A  =  18°  12'. 
2  6c      2  X  7  X  10    140 

cos^  =  ^'^  +  ^'-^'  =  ^QQ  +  ^^-^^^g^  =  .8375.     ...^  =  33°  7' 30". 
2  ca      2  X  10  X  4    80 

cos  o^  «^  +  &^  -  c^  ^  16  +  49  -  100  ^ -35  ^  _ ^^^^      .^  0  =  128°40'52". 
2x4x7  56 

Angle  C  is  in  the  second  quadrant  since  its  cosine 
is  negative. 

Check:  18°12'+33°7'30"  +  128°40'52"=180°0'22"., 
The  discrepancy  is  due  to  the  fact  that  four-place 
tables  were  used  in  the  computation.  Had  five-place 
tables  been  used,  the  discrepancy  would  have  been  less. 

2.  In  PQB,  p=    9,  g  =  24,  r  =  27.     Find  P,  Q,  B. 

3.  In  RS2\    r  =  21,  s  =  24,  «  =  27.     Find  B,  S,  T. 

4.  In  ABC,  a  =  12,  &  =  20,  c  =  28.  Find  A,  B,  C. 
6.  In  ABC,  a  =  80,  &  =  26,  c  =  74.  Find  A,  B,  G. 
6.  Solve  Ex.  1,  using  five-place  tables. 

59.  The  aid  of  logarithms  in  the  solution  of  triangles.     It  was 

pointed  out  in  Art.  6  that  an  expression  is  adapted  for  logarithmic 
computation  when,  and  only  when,  it  is  decomposed  into  factors. 
In  Cases  I.,  II.,  Arts.  55,  56,  the  expressions  used  in  solving  the 
triangle  can  be  computed  with  the  help  of  logarithms.  On  the 
other  hand,  the  side  opposite  to  the  given  angle  in  Case  III., 
Art.  57,  and  the  angles  in  Case  IV.,  Art.  58,  are  found  by  evalu- 
ating expressions  which  are  not  adapted  to  the  use  of  logarithms. 
Other  relations  between  the  sides  and  angles  of  a  triangle  will  be 
found  in  Arts.  61,  62.  By  these  relations  the  computations  in 
Cases  III.,  IV.,  can  be  made  both  without  and  with  logarithms. 
These  relations  are  useful  not  merely  for  purposes  of  computation ; 
they  are  important  in  themselves,  and  valuable  because  many  im- 
portant properties  of  triangles  can  be  deduced  from  them. 

The  explanations  given  in  Arts.  55-57  are  presupposed  in  Arts. 
60-62.  The  general  directions  to  be  observed  in  working  the 
problems  are  as  follows: 


59,60.] 


USE  OF  LOGARITHMS  IN   CASES  /.,   //. 


107 


1.  Write  down  all  the  formulas  wliicli  will  he  used  in  the  compu- 
tation. 

2.  Express  these  formulas  in  the  logarithmic  form. 

[As  soon  as  the  student  perceives  that  this  step  does  not  afford 
any  additional  assistance,  it  may  be  omitted.  See  Art.  27,  Ex.  1^ 
Note  6.] 

3.  Mahe  a  skeleton  scheme,  and  arrange  the  arithmetical  work 
neatly  and  clearly. 

The  skeleton  schemes  in  the  worked  examples  that  follow,  are 
apparent  when  the  numbers  are  omitted.* 

Checks:  The  various  formulas  can  serve  as  checks  on  the  re- 
sults of  one  another.  The  relations  derived  in  Exs.  1,  2,  Art. 
54  a,  are  also  useful  as  checks. 

60.  The  use  of  logarithms  in  Cases  I.,  II.  An  example  worked 
out,  will  give  suf&cient  explanation. 


EXAMPLES. 

1.  In  ABC,    given :  a  =  447,      To  find  :  B  = 

b  =  576,  C  = 

A  =  47°  35'.  c  = 


(Write  the  results  here.) 


Since  a  <  &,  there  may  be  two  sohitions.     Construc- 
tion shows  there  are  two  sohitions. 


Formulas : 


sin  ABC  = -sin  A  =  sin  ABiC. 

a 


ACB=180°-{A+ABC).    ACBi  =  lSO°-(A-]-ABi^C). 


AB 


a 


sin^ 


sin  ACB. 


ABi 


a 


sin^ 


sin^C^i. 


.*.  log  sin  ABC  =  log  &  +  log  sin  ^  —  log  a  =  log  sin  ABi  C ; 
log  AB  =  log  a  -f  log  sin  ACB  —  log  sin  A  ; 
log  ABi  =  log  a  +  log  sinACBi  —  log  sin  A. 


*  Cologarithms  are  not  used  in  the  solutions  in  the  text.  In  extensive 
computations  the  use  of  cologarithms  is  favoured  by  many  computers  ;  but  it 
seems  best  for  beginners  in  trigonometry  first  to  become  accustomed  to  the 
obvious  and  direct  method  of  working  with  logarithms. 


108 


PLANE  TRIGONOMETRY, 


[Ch.  VIL 


log  a  =  2.65031 

log  6  =2.76042 

logsin^  =  9.86821  -10 


•.  log  sin  B  =  9.97832  -  10 


.-.  ABC  =  72°  2'  45" 
.-.  ACB  =  60°  22' W 
log  sin  ACB  =  9.93914  -  10 
.-.  log^^  =  2.72124 
.-.  AB  =  526.3 


and  ^5iC=107°57'15" 
.-.  ACBi  =  24^  2T  i'o" 
log  sin  ACBi  =  9.61710  -  10 
.-.  log  ^i?i  =  2.39920 
.-.  ^i;i  =  250.7 


In  obtaining  log  J.^,  for  instance,  log  sin  ^C^  may  be  written  on  the 
margin  of  a  slip  of  paper,  placed  under  log  a,  the  addition  made,  log  sin  A 
placed  beneath,  and  the  subtraction  made. 

Solve  the  triangle  ABC,  when  the  following  elements  are  given  : 
2.   A  =  63°  48',  B  =  49°  25',  a  =  825  ft. 
Z.   B  =  128°  3'  49",  O  =  33°  34'  47",  a  =  240  ft. 

4.  ^  =  78°  30',  b  =  137  ft.,  a  =  65  ft. 

5.  a  =  275.48,  b  =  350.55,  B  =  60°  0'  32". 

6.  c  =  690,  a  =  464,  A  =  37°  20'. 

7.  a  =  690,  &  =  1390,  A  =  21°  14'  25". 


61.  Relation  between  the  sum  and  difference  of  any  two  sides  of  a 
triangle.  The  Law  of  Tangents.  Use  of  logarithms  in  Case  III. 
In  any  triangle  ABC,  for  any  two  sides,  say  a,  6, 


a  _  sin  A 
b      sin  B 

b      sin  ^  —  sin  B 


a  -f-  6      sin  A  +  sin  B 


[By  equation  (1),  Art.  54.] 
[By  composition  and  division.] 


^pos^A±B)m^^A-:B)^  [Art. 52, Formulas (5), (6).] 
2sm:L(^A  +  B)cos^(A-B)    ^  '  \  J>\  J  ^ 


«  +  ^     tanliA  +  B) 


[Art.  44,  A,  B.]         (1) 


That  is,  the  difference  of  any  trvo  sides  of  a  triangle  is  to  their 
sum  as  the  tangent  of  half  the  difference  of  their  opposite  angles  is 


61.] 


LAW  OF  TANGENTS,    CASE  III, 


109 


to  the  tangent  of  half  their  sum.     This  is  sometimes  called  the 
law  of  tangents. 


Kow  A-\-B=lSO°-G,  and,  consequently,  i{A-\-B)=90° - 
C 


O 


Hence,  tan  ^(A  +  B)  =  cot  — ,  and,  accordingly,  relation  (1)  may 
be  written 


tan^(JL-B)  =  |^cot|C. 


2 


(2) 


Formulas  for  b,  c,  and  c,  a,  similar  to  the  formulas  for  a,  b  in 
(1),  (2),  can  be  derived  in  the  same  way  as  (1),  (2),  have  been 
derived.  These  formulas  can  also  be  written  down  immediately, 
on  noticing  the  symmetry  in  formulas  (1),  (2). 

Ex.  Write  the  formulas  for  sides  b,  c  and  c,  a.  Derive  these 
formulas. 

Case  III,  In  a  triangle  ABC^  a,  b,  (7,  are  known,  and  c,  B,  Ay 
are  required.  Here,  ^  (A-{-  B)=  90°  —  ^Cj  also,  ^(A  —  B)  can  be 
found  by  (2).     Hence,  A  and  B  can  be  found ;  for 

J.  =  |(J.  +  J5)  +  i(J.-^),  and  B  =  ^(A  +  B)~-L{A-B). 

The  side  c  can  then  be  found  by  (1),  Art.  54.  (In  using  (1),  (2), 
write  the  greater  side  and  the  greater  angle  first,  in  order  that 
the  difference  may  be  positive.)  Fonnulas  (1),  (2),  can  also  be 
used  as  a  check  in  the  cases  discussed  in  the  preceding  aii:icles. 
Other  checks  will  be  shown  in  the  next  article. 


1.   In  triangle  ABC,  Given 


EXAMPLES. 

h  =  472,       Find  :  B  = 
c  =  324,  C  = 

^  =  78°40^.  a  = 


Formulas :  tan  ^  (-B  —  C)  = 


cot  I  A. 


b  +  c 

B  =  l{B+C)+KB-  C). 
C=KB+  C)-l(B-  C). 


h  sin  A 


csin^ 


Checks :  A  +  B+  C  = 
shown  in  ihe  next  article. 


sin  B  sin  C 

180°,  formulas  in  preceding  articles,  and  formulas 


110  PLANE  TRIGONOMETRY,  [Ch.  VII. 

log  tan  KB  -  O)  =  log(6  -  c)  +  log  cot  ^  ^  -  log(6  +  c), 

log  a  =  log  &  +  log  sin  A  —  log  sin  B ;  or  =  log  c  +  log  sin  A 
—  log  sin  C. 

h  =  472  log(6  -  c)  =    2. 17026  log  6  =  2.67394 

c  =  324  log(&  +  c)=    2.90091  log  sin  ^  =  9.99145  - 10 

A  =  78°  40'  log  cot  ^  ^  =  10.08647-10    log  sin  B  =  9.95223  -  10 

6-c  =  148         .-.  log  tan  K^-C7)=   9.35582-10      .-.  log  a  =  2.71316 
6  +  c  =  796  .-.  i^B-  C')=  12°  47'  1"  .-.  a  =  516.6 

I  ^  =  39°  20'  KB+  Q  =  50°  40' 

.-.  ^  =  63°  27'  1" 
.-.   C=  37°  52' 59" 

Check :  ^  +  5  +  C  =  78°  40'  +  63°  27'  1"  +  37°  52'  59"  =  180°. 

Note.     Formulas  (1),  (2),  are  adapted  to  logarithmic  computation;  but 
the  computations  can  be  made  without  the  aid  of  logarithms. 

2.  Solve  ABC,  given  b  =  352,  a  =  266,  C  =  73°. 

3.  Solve  PQR,  given  j?  =  91.7,  q  =  31.2,  R  =  33°  7'  9". 

4.  Solve  ABC,  given  a  =  960,  b  =  720,  O  =  25°  40'. 

5.  Solve  ABC,  given  b  =  9.081,  c  =  3.0545,  Jl  =  68°  14'  24". 

6.  Solve  Exs.  1,  5,  Art.  57,  using  the  formulas  of  this  article,  without 
logarithms. 

62.   Trigonometric  ratios  of  the  half  angles  of  a  triangle.     Use  of 
logarithms  in  Case  IV.     In  any  triangle  ABC, 

cos  A  =  ^'+J^'-<  [Art.  54  (4).] 

Now  l-cos^  =  2sin2i^,  \ 

and  1  +  cos  ^  =  2  cos^  \  A.       [Art.  50  (7),  (8).] 

Also.l-cos^=l-^^^+^^-^^  =  ^^^-(^^  +  ^^-^ 
'  2  6c  2  he 

_ a^-(b^  +  c^- 2  be)  ^a'-jb-  cf 
2  be  2  be 

2  6c 
.'.  2sin'»^^=:(^-^  +  ^)(^+^-")  (1) 


62.]                    USE  OF  LOGARITHMS  IN  CASE  IV.  Ill 

Also,l  +  cos^  =  l+       ^^^       = ^26^ ^ 

_  (6  +  cf  -  g^ 
2  6c 

.,  2cos^A  =  (^±^±f^^±^^^^  (2) 

Let  a4-&  +  c  =  2s; 

then  2(s  —  c)  =  (a  +  6  +  c)  —  2c  =  a  +  &  —  a 

Similarly  2  (s  —  6)  =  a  —  &  +  c, 

2  (s  —  a)  =  —  a  +  6  +  c. 

The  substitution  of  these  values  in  (1)  and  (2)  gives 

o   .   21    .      2(s-b)'2(s  —  c)      o«    21  A      2s'2(s-a) 
2sm^A  =  ^ br '^    2cosH^  = ^^. 

.«.  sm4^  =  ^^-^^^^-^);  (3) 

cos4^=5l^^.  (4) 
Since  tam'  J  ^  =  sin'  ^  ^  -s-  cos'  |  A,  it  follows  that 

ten''U  =  ^''-^H^-">-  (6) 


sm^ 


2  \      s(8-a) 


Note.  By  geometry,  b  +  c>a.  Hence,  — a  +  &  +  c>0,  and,  accord- 
ingly, s  —  a  is  positive.  Similarly,  s  —  b,  s  —  c,  are  also  positive.  Therefore, 
the  quantities  under  the  radical  signs  are  positive.  The  positive  sign  must 
be  given  to  the  radical,  for  A  is  less  than  180°,  and  consequently  J  A  lies 
between  0°  and  90° 


112  PLANE  TRIGONOMETRY.  [Ch.  VII. 

Similar  formulas  hold  for  i  B  and  \  C.  They  can  be  deduced 
in  the  same  manner  as  those  for  i  A ;  or,  they  can  be  written 
immediately,  from  the  symmetry  apparent  in  the  formulas  (3)-(5). 
The  student  is  advised  to  derive  the  similar  formulas  for  ^  B, 
i  O,  viz. : 

sin2i5  =  (£j=i^K£j=^;  cosH^  =  ^^^-^=-^;     (3') 

sinHC7=^'"^^^'~^^-  cosHO  =  '^'~'^'      (4') 

^  ah  ^  ah  ^   ^ 


^  s(s-b)      ' 

tanHO=^'"'^^^'~^>» 
^  s(s-c) 


(5') 


Formula  (5)  can  be  given  a  more  symmetrical  form;  for,  on 
multiplying  and  dividing  its  second  member  by  (s  —  a), 

^  s(s-ay  ' 


whence  tan ^  4  =  -1-^  /(8-a)(s-&)(g-c). 

2         s-a\  s 


If  y.^l(s-a)(8-b)(8-c)^ 


(6) 


(7) 


then  t&nlA^—^ —  (S) 

2         s-a  "-  ^ 

Similarly,        tan  45  =  —^,    tan  4  (7=-^.  (8') 

s  —  6  s  —  c 

When  all  the  sides  are  known,  the  angles  can  be  found  by 
means  of  formulas  (3)-(5')  or,  by  (7)-(8').  When  all  the  angles 
are  required,,  the  tangent  formulas  are  better,  since  fewer  loga- 
rithms are  required  than  in  (3),  (4),  (3'),  (4').  It  will  be  shown  in 
Art.  69  that  r  is  the  radius  of  the  circle  inscribed  in  the  triangle. 


63.] 


HEIGHTS  AND  DISTANCES. 


113 


1.    In  triangle  ABC,  a 
A,  B,  C. 

Formulas : 


EXAMPLES. 

25.17,6  =  34.06,0  =  22.17.  Find 


r=V^ 


a)(s  —  b)(s  —  c 


tan  ^A  = 


tan  i  C 


;  lau  f  o  = 

a  s  —  0  s  —  c 

:.  log r  =  i  [log  (s  -  a)  +  log  (s  -  6)  +  log  (s  -  c)  -  log s]. 

log  tan ^  J.  =  log r  -  log (s  -  a);   log  tan  | ^  =  log r 

log  tan  I  O  =  log  r  —  log  {s  —  c). 


A    C -^  22.17  B 

Fia.  60. 
log(s-6); 


Check : 

a  =  25.17 
b  =  34.06 
c  =  22.17 


2s  =  81.40 

s  =  40.70 

-a  =15.53 

-b=    6.64 

-  c  =  18.53 

Check : 


A  +  B-i-  C  =  180^ 

logs  =  1.60959 
log(s-a)  =1.19117 
log  (s- ft)  =0.82217 
log  (s-c)  =1.26788 

.-.  logr2  =  1.67163 

.-.  logr  =  0.83582 


log  tan  ^^=  9.64465-10 
i^  =  23°48'28" 

log  tan  1  B  =  10.01365  -  10 
^  5  =  45°  54' 

log  tan  10=  9.56794-10 
^C  =  20°  17'  35" 


.-.  A  =  47°  36'  56",   B  =  91°  48',    C  =  40°  35'  10" 
A  +  B  +  C=  180°  0'  6". 

2.  Solve  ABC,  given  a  =  260,  b  =  280,  c  =  300. 

3.  Solve  ABC  when  a  =  26.19,  b  =  28.31,  c  =  46.92. 

4.  Solve  PQB,  given  p  =  650,  q  =  736,  r  =  914. 

5.  Solve  EST,  given  r  =  1152,  s  =  2016,  t  =  2592. 

6.  Solve  Exs.  1,  4,  Art.  58,  using  formulas  (3)-(8'),  without  logarithms. 

63.  Problems  in  heights  and  distances.  Some  problems  in 
heights  and  distances  have  been  solved  in  Art.  29  by  the  aid  of 
right-angled  triangles.  Additional  problems  of  the  same  kind 
will  now  be  given,  in  the  solution  of  which  oblique-angled  tri- 
angles may  be  used.  It  is  advisable  to  draw  the  figures  neatly 
and  accurately.     The  graphical  method  should  also  be  employed. 

EXAMPLES. 

1.    Another  solution  of  Ex.  2,  Art.  29. 

In  the  triangle  ABP  (Fig.  23),  AB  =  100  ft.,  5^P=  30°,  PBA  =  180° 
—  45°  =  135°.  Hence  the  triangle  can  be  solved,  and  BP  can  be  found. 
When  ^P  shall  have  been  found,  then  in  the  triangle  CBP,  PPis  known  and 
£P  =  45°  ;  hence  CP  can  be  found.     The  computation  is  left  to  the  student. 


114  PLANE   TRIGONOMETRY,  [Ch.  VII. 

2.  Another  solution  of  Ex.  3,  Art.  29.  In  the  triangle  CBP  (Fig.  24), 
BP  =  30  ft. ,  BGP  =  40°  20'  -  38°  20^  =  2°,  PBC  =  90°  +  ZC^  =  128°  20'. 
Hence  CBP  can  be  solved  and  the  length  of  CB  can  be  found.  When  CB 
shall  have  been  found,  then,  in  the  triangle  LCB^  angle  0  =  38° 20',  CB  is 
known,  and  hence  LB  can  be  found.    The  computation  is  left  to  the  student. 

3.  Find  the  distance  betw^een  two  objects  that  are  invisible  from  each  other 
on  account  of  a  wood,  their  distances  from  a  station  at  which  they  are  visible 
being  441  and  504  yd.,  and  the  angle  at  the  station  subtended  by  the  distance 
of  the  objects  being  55°  40'. 

4.  The  distance  of  a  station  from  two  objects  situated  at  opposite  sides  of 
a  hill  are  1128  and  936  yd.,  and  the  angle  subtended  at  the  station  by  their 
distance,  is  64°  28'.     What  is  their  distance  ? 

5.  Find  the  distance  between  a  tree  and  a  house  on  opposite  sides  of  a 
river,  a  base  of  330  yd.  being  measured  from  the  tree  to  another  station,  and 
the  angles  at  the  tree  and  the  station  formed  by  the  base  line  and  lines  in  the 
direction  of  the  house  being  73°  15'  and  68°  2',  respectively.  Also  find  the 
distance  between  the  station  and  the  house. 

6.  Find  the  height  of  a  tower  on  the  opposite  side  of  a  river,  when  a 
horizontal  line  in  the  same  level  with  the  base  and  in  the  same  vertical  plane 
with  the  top  is  measured  and  found  to  be  170  ft.,  and  the  angles  of  elevation 
of  the  top  of  the  tower  at  the  extremities  of  the  line  are  32°  and  58°,  the 
height  of  the  observer's  eye  being  5  ft. 

7.  Find  the  height  of  a  tower  on  top  of  a  hill,  when  a  horizontal  base  line 
on  a  level  with  the  foot  of  the  hill  and  in  the  same  vertical  plane  with  the 
top  of  the  tower  is  measured  and  found  to  be  460  ft.  ;  and  at  the  end  of  the 
line  nearer  the  hill  the  angles  of  elevation  of  the  top  and  foot  of  the  tower 
are  36°  24',  24°  36',  and  at  the  other  end  the  angle  of  elevation  of  the  top  of 
the  tower  is  16°  40'. 

8.  A  church  is  at  the  top  of  a  straight  street  having  an  inclination  of 
14°  10'  to  the  horizon ;  a  straight  line  100  ft.  in  length  is  measured  along  the 
street  in  the  direction  of  the  church  ;  at  the  extremities  of  this  line  the  angles 
of  elevation  of  the  top  of  the  steeple  are  40°  30',  58°  20'.  Find  the  height  of 
the  steeple. 

-^  9.  The  distance  between  the  houses  C,  Z>,  on  the  right  bank  of  a  river 
and  invisible  from  each  other,  is  required.  A  straight  line  AB^  300  yd.  long, 
is  measured  on  the  left  bank  of  the  river,  and  angular  measurements  are 
taken  as  follows  :  ^50=  53°  30',  CBD=^5°  15',  CAD=ii7°,  i>^5=58°20'. 
What  is  the  length  CD  ? 

10.  A  tower  CD,  C  being  the  base,  stands  in  a  horizontal  plane  ;  a  hori- 
zontal line  AB  on  the  same  level  with  the  base  is  measured  and  found  to  be 
468  ft.;  the  horizontal  angles  ^^C,  ABC,  are  equal  to  125° 40',  12° 35', 
respectively,  and  the  vertical  angles  CAD,  CZ?Z>,  are  equal  to  30°  20',  11°  50', 
respectively.    Find  the  height  of  the  tower  and  its  distances  from  A  and  B. 


64.]  SUMMABY.  115 

11.  A  base  line  AB  850  ft.  long  is  measured  along  the  straight  bank  of  a 
river;  O  is  an  object  on  the  opposite  bank;  the  angles  BAC,  ABC,  are 
observed  to  be  63°  40',  37°  15',  respectively.     Find  the  breadth  of  the  river. 

12.  A  tower  subtends  an  angle  a  at  a  point  on  the  same  level  as  the  foot 
of  the  tower  and,  at  a  second  point,  h  feet  above  the  first,  the  depression  of 
the  foot  of  the  tower  is  j3.     Show  that  the  height  of  the  tower  is  h  tan  a  cot  ^. 

13.  The  elevation  of  a  steeple  at  a  place  due  south  of  it  is  45°,  and  at 
another  place  due  west  of  it  the  elevation  is  15°.  If  the  distance  between 
the  two  places  be  a,  prove  that  the  height  of  the  steeple  is  a(\/3  — 1)  ^2^2. 

14.  The  elevation  of  the  summit  of  a  hill  from  a  station  ^  is  a  ;  after 
walking  c  feet  toward  the  summit  up  a  slope  inclined  at  an  angle  ^  to  the 
horizon  the  elevation  is  7.  Show  that  the  height  of  the  hill  above  A  is 
c  sin  a  sin  (7  —  jS)  cosec  (7  —  a)  ft. 

64.  Summary.  The  preceding  discussions  on  the  solution  of 
triangles  have  shown  that  a  triangle  may  be  solved  in  the 
following  ways : 

I.  By  the  graphical  method.     [Arts.  10,  14,  21-24.] 

II.  If  the  triangle  is  right-angled,  it  can  be  solved,  either  with 
or  without  logarithms,  by  the  methods  shown  in  Arts.  25-27. 

III.  If  the  triangle  is  oblique,  it  can  be  divided  into  right- 
angled  triangles,  each  of  which  can  be  solved  by  either  of  the 
methods  II.     [Art.  34.] 

IV.  The  triangle,  whether  right-angled  or  oblique,  can  be  solved 
without  using  logarithms,  by  means  of  formulas  (1),  (3),  Art.  54; 
(1)  or  (2),  Art.  61 ;  (3)-(8),  Art.  62. 

V.  The  triangle,  whether  right-angled  or  oblique,  can  be  solved 
with  the  use  of  logarithms,  by  means  of  formulas  (1),  Art.  54; 
(1)  or  (2),  Art.  61 ;  (3)-(8),  Art.  62. 

Checks :  Any  formula  not  employed  in  the  computation  can  be 
employed  as  a  check ;  that  is,  as  a  test  for  the  correctness  of  the 
result. 

Two  things  are  necessary  on  the  part  of  one  who  wishes  to  do 
well  in  the  solution  of  triangles : 

(1)  The  formulas  referred  to  above  should  be  clearly  under- 
stood and  readily  derived. 

(2)  The  arithmetical  work  required  should  be  done  accurately, 

N.B.  Questions  and  exercises  suitable  for  practice  and  revieuo  on  the 
subject-matter  of  this  chapter  will  be  found  at  pages  189-193. 


CHAPTER  VIII. 


SIDE    AND    AREA    OF    A    TRIANGLE.      CIRCLES    CON- 
NECTED WITH   A  TRIANGLfe-^-':^'" 

65.  Length  of  a  side  of  a  triangle  in  terms  of  the  adjacent  sides 
and  the  adjacent  angles.  In  this  proof,  regard  is  paid  to  the  con- 
ventions about  signs,  described  in  Arts.  36,  37.  Let  ABC  be  any 
triangle.  From  A  draw  AD  perpendicular  to  BC,  or  BG  pro- 
duced. The  positive  direction  of  BC  is  in  the  direction  of  V. 
[At  the  first  reading,  only  Fig.  61  a  may  be  regarded.] 


B  DC 

Fig.  61a. 


V 


B  C 

Fig.  61a 


BO  =  BD-hDC, 
=  -GD  +  BD, 

=  -CA  cos  VGA  +  BA  cos  VBA.     [Art.  40.] 
But  VGA  =  180°  -  AGB. 

.-.  cos  VGA  =  cos  (180°  -  AGB)  =  -  cos  AGB.   [Art.  45.] 
.-.  BG  =  GA  cos  AGB  +  BA  cos  GBA ; 
i.e.  a  =  6  cos  C  +  c  cos  B,  (1) 

Therefore^  in  any  triangle  each  side  is  equal  to  the  sum  of  the 
products  of  each  of  the  other  sides  by  the  cosine  of  the  angle  which  it 
makes  with  the  first  side. 

When  (7  is  a  right  angle,  (1)  reduces  to  a  =  c  cos  B. 

Example.  Write  the  corresponding  formulas  for  b  and  c.  Derive  these 
formulas. 

116 


65,  66.] 


AREA   OF  A   TRIANGLE. 


117 


66.   Area  of  a  triangle.     Suppose  that  the  area  of  a  triangle 
ABC  is  required.     Let  the  length  of  the  perpendicular  DC  from 


A  D      B 

Fig.  63a, 


C  to  AB,  «or  AB  produced,  be  denoted  by  p,  and  let  the  area  be 
denoted  by  S.     The  following  cases  may  occur  : 

I.  One  side  and  the  perpendicular  on  it  from  the  opposite  angle 
known  J  say  (c,  p). 

S  =  \cp.     [By  geometry.]   ^  (1) 

II.  Two  sides  and  their  included  angle  Jcnoiun,  say,  6,  c,  A.     (See 
Figs.  62  a,  62  b.) 

S=:^cp  =  iC' AC  sin  BAO.  [Art.  40.] 

.  •.  S  =  lbc  sin  A.     [Compare  Art.  31.]     (2) 

III.  Three  sides  known. 

/S  =  ^  6c  sin  ^  =  i  &c  •  2  sin  1  ^  cos  I  A, 


be 


(s-b)(s-c)     Ri^ 
\  ha  \       ho.       ' 


be  ^      be 

.'.  S  =  Vs{s  —  a) (s  —  b) (s  -  c), 


[Art.  50.] 
[Art.  62.-] 
(3) 


That  is,  the  area  of  a  triangle  is  equal  to  the  square  root  of  the 
product  of  half  the  sum  of  the  sides  by  the  three  factors  formed  by 
subtracting  each  side  in  turn  from  this  half  sum.  See  Art.  34  a 
for  another  derivation  qf  this  formula. 

IV.    One  side  and  the  angles  known,  say,  a,  A,  B,  G. 

a  sin  JB 


^  ab  sin  O. 


S  = 


Now  b  = 
1  a^  sin  O  sin  B 


sin  J. 


2         sin^ 
Example.     Write  and  also  derive  the  similar  formulas  in  b  and  c. 


(4) 


118  PLANE  TRIGONOMETRY,  [Ch.  VIIL 

EXAMPLES. 

1.  Find  the  areas  of  the  triangles  in  Exs.  1-5,  Art.  61. 

2.  Find  the  areas  of  the  triangles  in  Exs.  1-5,  Art.  62. 

3.  Find  the  areas  of  the  triangles  in  Exs.  2,  3,  Art.  60. 

67.  Area  of  a  quadrilateral  in  terms  of  its  diagonals  and  their 
angle  of  intersection. 

Area^J5C'i)=area^i)0  +  area  ABC.  %^ 

Area  AD(7=area  ALB  +  area  QLB 

=i  AL  '  LD  sin  ALD 

A  B 

+i  GL  '  LD  sin  CLD  [Art.  m  (2).]  fig.  63. 

=^{AL+LC)DL  smALD,  (since  sin  CLD=smALD) 

=^  AG '  DL  sin  ALD. 

Similarly, 

area  ABG  =^  AG  -  BL  sin  BLG=  ^  AG  -  BL  sin  ALD. 

.-.  area  ABGD  =  i  AG(DL+LB)  sin  DLA=^  AG  -  BD  sin  DLA. 

.'.  area  of  a  quadrilateral  is  equal  to  one-half  the  product  of 
the  two  diagonals  and  their  angle  of  intersection. 

EXAMPLES.  ^ 

1.  Find  the  area  of  a  quadrilateral  whose  diagonals  are  108,  240  ft.  long, 
and  inclined  to  each  other  at  an  angle  67°  40'.  Find  the  sides  and  angles 
of  a  parallelogram  having  these  diagonals. 

2.  So  also  when  the  diagonals  are  360,  570  ft.  long,  and  their  inclination 
is  39°  47'. 

3.  The  diagonals  of  a  parallelogram  are  347  and  264  ft.,  and  its  area 
is  40,437  sq.  ft.     Find  its  sides  and  angles. 

4.  Solve  an  isosceles  trapezoid,  knowing  the  parallel  sides  a =682. 7  metres, 
c  =  1242.6  metres,  and  the  non-parallel  equal  sides  b  =  d  =  986.4  metres. 
Find  the  angles,  the  area,  the  lengths  and  angle  of  inclination  of  the 
diagonals. 

___._.._ _ __ __.^J 


67-69.]       CIRCLES   CONNECTED    WITH  A    TRIANGLE. 


119 


68.  The  circumscribing  circle.  Let  the  radius  of  the  circle 
described  about  a  triangle  ABC  be  denoted  by  R.  It  has  been 
shown  (see  equation  (2),  Art.  54)  that 

2  sin  ^     2  sin  B     2  sin  C  ^  ^ 

That  is,  the  radius  of  the  circumscribing  circle  of  aiiy  triangle  is 
equal  to  half  the  quotient  of  any  side  by  the  sine  of  the  opposite 
aiigle. 

From  (2),  Art.  m,  sin^ 
first  of  equations  (1),  gives 


2S_ 
be  ' 


Substitution  of  this  in  the 


It  = 


abc 


69.  The  inscribed  circle.  Let  the 
radius  of  the  circle  inscribed  in  a  tri- 
angle ABC  be  denoted  by  r.  Join 
the  centre  0  and  the  points  of  contact 
L,  M,  N.  By  geometry,  the  angles 
at  L,  M,  ]^  are  right  angles.  Draw 
OA,  OB,  OC 

Area  500+ area  00^ 

+  area  AOB  =  area  ABC. 


(2) 


iar-^i 


br  +  i 


I.e. 


cr  =  V  s(s 
sr=:S. 


i){s  —  b)(s  —  c),  or  S. 


.-.  r=\ 


(s  -  a)(s  -  b)(s 


s 


(3) 


That  is,  the  length  of  the  radius  of  the  inscribed  circle  of  a  triangle 
is  equal  to  the  number  of  units  in  its  area  divided  by  half  the  sum 
of  the  lengths  of  its  sides.     See  reference  in  Art.  62. 

Note.  Formula  (8),  Art.  62,  can  be  readily  derived  from  Fig.  64.  By 
geometry,  AN  =  MA,  BL  =  NB,   CM  =  LC. 

NO 


Now 

But  NO  =  r,  and 

AN  =  (iAN 


tan  ^  A  =  tan  BA  0  = 


AN 


BL+  CM)-{BL-VLC)=s 
.'.  tan  \  A= . 


120 


PLANE  TEtGONOMETRY, 


[Ch.  VIIL 


I.e. 


70.  The  escribed  circles.  An  escribed  circle  of  a  triangle  is 
a  circle  that  touclies  one  of  the  sides 
of  the  triangle  and  the  other  two  sides 
produced. 

Let  r^  denote  the  radius  of  the 
escribed  circle  touching  the  side  BC 
opposite  to  the  angle  A.  Join  the 
centre  Q  and  the  points  of  contact 
L,  M,  N.  By  geometry,  the  angles 
at  L,  My  iV  are  right  angles.     Draw 

QA,  QB,  qa 

Area  ABQ  +  area  CAQ  —  area  BCQ  =  area  ABO. 
'-'  iKG  +  irJ)-iraa  =  S, 
...  ^(c  +  b-a)r,  =  S; 
(s  —  d)r^  =  S, 
S 


ra  = 


Similarly, 


.  S 


8  — a 

S 

S—  G 


Other  interesting  relations  between  the  sides,  angles,  and  related 
circles,  of  a  triangle,  are  indicated  in  the  exercises  in  the  latter  part 
of  the  book. 

EXAMPLES. 

1.  Find  the  radii  of  the  circumscribed,  inscribed,  and  escribed  circles  of 
some  of  the  triangles  in  Arts.  55-58. 

2.  Find  the  radii  of  these  related  circles  of  some  of  the  triangles  in 
Exs.  1-3,  Art.  66. 

N.B.  Questions  and  exercises  suitable  for  practice  and  review  on  the 
subject-matter  of  this  Chapter  will  he  found  at  pages  193,  194. 


CHAPTER  IX. 

RADIAN  MEASURE. 

71.  The  radian  defined.  The  system  of  measuring  angles  with 
a  degree  as  the  unit  angle,  was  described  in  Art.  11.  Since  the  time 
of  the  Babylonians  this  system  has  been  the  common  practical 
method  employed.  Another  method  of  measuring  angles  was 
introduced  early  in  the  last  century.  This  method  is  used  to 
some  extent  in  practical  work,  and  is  universally  used  in  the 
higher  branches  of  mathematics.     It  is  employed,  on  account  of 


Fig.  66. 


its  great  convenience,  in  the  larger  and  more  important  part  of 
what  is  now  called  trigonometry,  namely,  the  part  which  is  not 
concerned  with  the  measurement  of  lines  and  angles,  but  which 
pursues  investigation  of  the  properties  of  the  quantities  that,  so 
far  in  this  book,  have  been  called  the  trigonometric  ratios.  A 
very  little  knowledge  of  the  trigonometric  ratios  is  sufficient  for 
the  solution  of  triangles.  The  more  detailed  and  extended  study 
of  angles  and  their  six  related  numbers,  constitutes  part  of  what 
is  sometimes  called  Higher  Trigonometry,  but,  more  generally. 
Analytical  Trigonometry.  This  subject  is  a  large  one,  and  has  close 
connections  with  many  other  branches  of  modern  mathematics. 

121 


122  PLANE  TRIGONOMEmr,  [Ch.  IX. 

The  system  of  angular  measurement  now  to  be  described,  is 
sometimes  referred  to  as  the  theoretical  system  of  measurement. 
In  this  system  the  unit  angle  is  the  angle  which  at  the  centre  of  a 
circle  subtends  an  arc  equal  in  length  to  the  radius.  This  unit 
angle  is  called  a  radian.  Thus,  if  a  circle  with  any  radius  be  de- 
scribed about  0  as  a  centre,  and  an  arc  AB  he  taken  equal  in 
length  to  the  radius,  then  the  angle  AOB  is  a  radian. 

72.  The  value  of  a  radian.  In  order  that  a  quantity  may  be 
used  as  a  unit  of  measurement,  it  must  have  a  fixed  value ;  that 
is,  using  the  customary  mathematical  phrase,  it  must  be  a  constant 
quantity.  The  proof  that  a  radian  has  a  fixed  value,  or  is  a  con- 
stant quantity,  depends  upon  two  geometrical  facts,  viz. : 

(a)  In  the  same  circle  two  angles  at  the  centre  are  in  the  same 
ratio  as  their  intercepted  arcs. 

(6)  The  ratio  of  a  circumference  of  a  circle  to  its  diameter  is 
the  same  for  all  circles.     [See  Art.  9  (6).] 

For  the  proof  of  (a),  reference  may  be  made  to  any  plane  geome- 
try ;  for  instance,  to  Euclid  VI.,  33.*  The  proof  of  (b)  is  not  con- 
tained in  all  geometries ;  for  instance,  Euclid  does  not  give  it.f 
Accordingly,  an  outline  of  such  a  proof  and  the  calculation  of  tt 
are  given  in  Note  C  of  the  Appendix.  This  note  should  now  be 
studied  by  those  whose  course  in  plane  geometry  has  not  included 


*  The  truth  of  theorem  (a)  can  easily,  by  an  inductive  method,  be  made 
evident  to  students  who  have  not  proved  the  theorem  in  plane  geometry. 
Thus,  on  taking  angles  which  are  twice,  three  times,  four  times,  one-half,  one- 
third,  etc.,  of  a  given  angle,  it  can  be  seen  that  their  respective  arcs  bear  the 
same  relations  to  one  another. 

t  Euclid  lived  about  323-283  b.c.  Archimedes  (287  ?-212  b.c),  the 
greatest  mathematician  of  antiquity,  measured  the  length  of  the  circle  and 
the  area  contained  by  it,  and  also  measured  the  surface  of  the  sphere.  He 
showed  that  the  ratio  of  the  circle  to  its  diameter  lies  between  ^^  and  -2^. 
In  1794  a  French  mathematician,  Adrien  Marie  Legendre  (1762-1833),  pub- 
lished his  Elements  of  Geometry,  in  which  the  works  of  Euclid  and  Archi- 
medes on  elementary  geometry  are  blended  together.  The  elementary  text- 
books now  in  use  on  the  continent  of  Europe  and  in  the  United  States,  are 
written  mainly  on  Legendrean  lines  ;  the  geometrical  text-books  generally 
studied  throughout  the  British  Empire,  are  editions  of  Euclid's  Elements. 


72,  73.]  THE  BADlAN.  123 

the  measurement  of  the  circle.    Theorems  (a)  and  (h)  are  assumed 
in  what  foHows. 

T    -c^-     oi'     tJi^  radian  AOB  arc  AB  rT>     /  n  n 

In  Fig.  60,    -J-T77 -. = r^ .    .    ,      [By  (a).] 

4  ri^/ii  angles        circumference  of  circle 

1  2 

.*.  the  radian  =  -—  x  4  right  angles  =  -  X  right  angle.      (1) 

Since  all  right  angles  are  equal,  and  since  each  radian  is  a  fixed 
2 
fraction,  namely,  -,  of  a  right  angle,  it  follows  that  all  radians 

IT 

are  equal.     It  will  be  remembered  that  the  unit  in  the  common 
practical  system  is  one-ninetieth  of  a  right  angle. 

From  (1), 

A  radian  =  i^°  (2) 

IT 

-  — 5:55! —  =  57°  IV  44 ".81  approximately  * 


3.14159... 
=  206265"  approximately. 

Ex.  "With  a  protractor  lay  off  an  angle  approximately  equal  to  a  radian. 
Compare  it  with  angle  60°.  An  angle  60°,  at  centre  of  a  circle,  is  subtended 
by  a  chord  equal  in  length  to  the  radius ;  a  radian  is  subtended  by  an  arc 
equal  in  length  to  the  radius. 

73.   The  radian  measure  of  an  angle.     Measure  of  a  circular  arc. 

The  radian  measure  of  an  angle  is  the  ratio  of  the  angle  to  a  radian. 
[See  Art.  8.]  For  instance,  if  an  angle  A  is  twice  a  radian,  then 
its  radian  measure  is  2 ;  if  an  angle  B  is  two-thirds  of  a  radian, 
then  its  radian  measure  is  J.     This  is  expressed  thus : 

A  =  2  radians  =  2'' ;     ^  =  |  radians  =  f.  (3) 

Here,  r  is  used  as  the  symbol  for  radians  just  as  °  is  used  as 
the  symbol  for  degrees  in  23°.      In  general  discussions  the  radian 

*  The  value  of  the  radian  has  been  calculated  by  J.  W.  L.  Glaisher  to  41 
places  of  decimals  of  a  second.  \_Proc.  Land.  Math.  Soc,  Vol.  IV.  (1871-73), 
pp.  308-312.] 


124  PLANE  TRIGONOMETRY.  [Ch.  IX. 

measure  of  an  angle  is  often  expressed  by  Greek  letters ;  thus, 
the  angles  a,  jS,  6,  cf),  etc.,  contain  a,  p,  6,  cjy,  etc.,  radians.  In 
these  cases  the  symbol  r  is  usually  omitted,  but  it  is  always 
understood  that  the  radian  is  the  unit  of  measurement. 

If  the  circular  arc  subtended  by  an  angle  is  equal  in  length  to 
twice  the  radius,  then  the  radian  measure  of  the  angle  is  obviously 
two;  if  the  arc  is  one-half  t\iQ  length  of  the  radius,  then  the  angle 
contains  half  a  radian.  Tlie  radian  measure  of  an  angle  may  he 
given  a  second  definition,  which  depends  on  Theorem  (a).  Art.  72. 
Let  AOP,  Fig.  QQ>,  be  any  angle,  and  AOB  be  a  radian.  Describe 
a  circle  with  any  radius  OA,  equal  to  r,  about  the  vertex  O  as 
a  centre.  Let  arc  AB  be  equal  to  the  radius,  and  draw  OB.  Then 
angle  AOB  is  a  radian,  by  the  definition  in  Art.  71. 

j^^le^OP^ar^    [Th.  (a).  Art.  72.]  (4) 

radian  AOB     arc  AB     ^        ^  ^'  ^  ^  ^ 

angle  AOP       arc  AP  zt.. 

I.e.  —-^ —  = —  (5) 

the  radian      the  radius 

That  is,  the  number  of  radians  in  an  angle,  or  the  radian  measure 
of  an  angle,  is  the  answer  to  the  question :  how  many  times  does  any 
circular  arc  subtended  by  it,  contain  the  radius  %  Thus,  for  example, 
the  radian  measures  of  the  angles  which  subtend  circular  arcs 
equal  in  length  to  2,  3,  1.5,  .825  radii  are  2,  3,  1.5,  .825,  respec- 
tively. 

The  circular  arc  subtended  by  360°  =  2  Trr ; 

2  ttT 

hence,  radian  measure  of  360°  = =  2  ir, 

r 

and  radian  measure  of  180°  =  tt.  (6) 

This  shows  that  an  angle  2  it  radians  is  described  each  time 
that  the  revolving  line  makes  a  complete  revolution.  Relation 
(6),  namely, 

180°  =  IT  radians,  (7) 

connects  the  two  systems  of  angular  measurement.  By  means 
of  (7),  an  angle  expressed  in  the  one  system  can  be  expressed 
in  the  other.  The  word  radians  is  usually  omitted  from  (7),  but 
is  always  understood.     Relation  (7)  may  also  be  deduced  directly 


73.]  RADIAN  MEASURE  OF  AN  ANGLE.  125 

from  (1).     Just  as  angles  are  considered  as  unlimited  in  mag- 
nitude, so  arcs  are  considered  as  unlimited  in  length. 

Note  1.     The  term  circular  measure  is  often  used  for  radian  measure, 


Ex.  1.   Express  30°  in  radian  measure. 
Since 


180»_..,   r-^;^; 

TT 

"6' 

Ako,30»--- 3.14159'... 

30°  =  T%^  =  77'     Also,  30°  =  ^  =  "-""7" =  .52359^  .... 

180  g  >  g  g 

Tlie  term  radians  and  the  symbol  for  radian  is  usually  omitted  from  the 
second  members  of  these  equations,  but  is  always  understood  to  be  there. 

Ex.  2.  Express  45°,  60°,  135°,  210°,  300°,  330°,  270°,  225°,  -  75°,  63°,  27°, 
—  33°,  —  150°,  in  radian  measure,  (a)  as  fractions  of  tt,  (&)  numerically,  on 
putting  TT  =  ^1^. 

Ex.  3.  Express  the  angle  ^^  tt  in  degrees. 
Here,  "t^tt"  means  "^^^^  x  -^i^  radians.''^ 
Since  tt  =  180°,   i^  tt  =  /^  x  180°  =  162°. 

Ex.  4.    Express  the  angles  — ,   -,   ^,   J,   J,   |7r,   ^tt,    10  tt,   4  7r,  Stt, 

2     3     4     6     5 

6  TT,  f  TT,  f  TT,  in  degrees,  and  their  complements  and  supplements,  in  radians. 
Ex.  6.   Express  the  angles  —  |  tt,  —  5  tt,  —  f  tt,  —  ii  tt,  —  25  tt,  in  degrees. 
Ex.  6.   Express  2''  (2  radians)  in  degrees. 
Since  tt  =  180°, 

TT 

.-.  2'-  =  2  X  —  =  114°  35'  29.6"  approximately. 

TT 

Ex.  7.   Express  — ,  4'',  3%  — ,  — ,  5'-,  lO**,  — ,  in  degrees. 
2  3     5  10 

Measure  of  an  arc.     Since,  by  (5), 

subtended  circular  arc  _  ^^^^^^^  ^^  ^^^i^,,^  j^  ^j^^  ^„glg_ 
radius 
then  len^h  of  arc  =  radius  x  number  of  radians  in  the  angle. 

If  a  denote  the  length  of  any  arc  AP,  r  the  radius,  6  the  radian 
measure  of  angle  AOPj  then 

a  =  r9.  (8) 


126  PLANE  TRIGONOMETRY,  [Ch.  IX. 

In  words  :  The  length  of  any  circular  arc  is  equal  to  the  product  of 
the  radius  and  the  radian  measure  of  its  subtended  central  angle.     For 

example,  the  arc  of  360°  =  2  tt  radii,  arc  of  180°  =  tt  radii,  etc. 
These  arcs  are  usually  referred  to  as  the  arcs  2  7r,  tt,  etc.;  but 
it  is  always  understood  that  the  radius  is  the  unit  of  measurement. 
The  symbol  tt,  which  always  denotes  the  incommensurable  num- 
ber 3.14159  •••,  can  thus  be  used  in  three  connections  in  trigo- 
nometry : 

(1)  With  other  numbers,  as  a  number  simply. 

(2)  With  reference  to  angles;  in  which  case  it  denotes  an  angle 
containing  tt  radians,  i.e.  3.14159  •••  radians. 

(3)  With  reference  to  arcs;  in  which  case  it  denotes  an  arc 
containing  3.14159  radii.  This  is  an  arc  subtended  by  a  central 
angle  of  tt  radians. 

The  expression  180°  =  tt  does  not  mean  180°  =  3.14159  ... ; 
it  means  180°  =  3.14159  radians. 

The  expression  ^^  arc  tt  "  does  not  mean  arc  3.1416 ;  it  means 
"arc  of  3.1416  radii.^^  In  any  particular  instance,  the  context 
will  show  to  what  tt  refers,  whether  to  angle  or  arc. 

It  is  evident  from  the  second  definition  of  radian  measure  that, 
like  the  trigonometric  ratios,  the  radian  measure  of  an  angle  is  also 
a  ratio  of  one  line  to  another,  namely,  the  ratio  of  the  subtended 
circular  arc  to  its  radius. 

Note  2.  If  the  radius  he  taken  as  unit  lengthy  then,  by  (8)  or  (5),  the 
number  of  units  of  length  in  the  arc  is  the  same  as  the  number  of  radians 
in  the  angle. 

EXAMPLES. 

8.  What  is  the  radian  measure  of  the  angle  which  at  the  centre  of  a 
circle  of  radius  1^  yd.  subtends  an  arc  of  8  in.?  Also  express  the  angle 
in  degrees. 

Let  6  denote  the  radian  measure  of  the  angle.     Then 


Since 


arc       8  in. 

_  8  _  4 

rad     1.5  yd. 

64     27' 

TT  =  180°, 

M^  =  l«^ 

••  (/?)'■  =  iV  X  22  X  180°  =  8°  29'  approximately. 


73.]  EXAMPLES.  127 

9.   Give  the  trigonometric  ratios  of 

TTTTTTTr  »  c  >,  e 

— ,    — ,    — ,    — ,    TT,    fTT,    —  #7r,    —  iTT,    —  #  T. 

6     4     3    2 

10.  Find  the  numerical  values  of  (a)  sin2  -  +  cos^  |  tt  +  tan2  — , 

6  3 

(6)  3sinf  Trcos-i^Trtan^^TT,  (c)  2  sin -2/ tt  cos -2/ tt  tan  ^/ tt. 

11.  Find  the  number  of  radians  (a)  as  fractions  of  tt,  (b)  numerically 
(on  putting  w  =  ^-^),  in  each  interior  and  exterior  angle  of  the  following  regu- 
lar polygons:  pentagon,  hexagon,  heptagon,  octagon,  decagon,  dodecagon, 
quindecagon. 

12.  Find  the  number  of  radians  and  the  number  of  degrees  in  the  follow- 
ing angles  subtended  at  the  centres  of  circles  :  (1)  arc  10  in.,  radius  3.5  in.; 
(2)  arc  }  ft.,  radius  2  ft.;  (3)  arc  1  mi.,  radius  7920  mi.;  (4)  arc  250  mi., 
radius  8000  mi.;  (5)  arc  10  yd.,  radius  10  mi.;  (6)  arc  ^  mi.,  radius  10  ft. 

13.  What  are  the  radii  when  an  arc  10  in.  in  length  subtends  central 
angles  containing  1,  2,  4,  6,  8, 12,  15,  20,  |,  ^,  f,  f ,  radians  respectively  ? 

14.  What  are  the  radii  when  an  arc  10  in.  in  length  subtends  central 
angles  containing  1°,  2°,  3°,  16^  28°,  120°,  30',  20',  10',  10",  20",  45",  re- 
spectively ? 

15.  In  a  circle  whose  radius  is  10  in.,  what  are  the  lengths  of  the  arcs 
subtended  by  central  angles  containing  1,  4,  7,  8,  12,  .5,  .375,  .125,  radians 
respectively  ? 

16.  In  the  circle  in  Ex.  15,  what  are  the  lengths  of  the  arcs  subtended  by 
central  angles  containing  2°,  25°,  48°,  135°,  250°,  30',  45',  30",  50",  respec- 
tively ? 

17.  What  are  the  areas  of  the  circular  sectors  in  Exs.  13,  15  ?  [See  Note 
C,  5.] 

N.B.  Questions  and  exercises  suitable  for  practice  and  review  on  the 
subject-matter  of  this  Chapter  will  be  found  at  pages  194,  195. 


CHAPTER  X. 

ANGLES  AND  TRIGONOMETRIC  FUNCTIONS. 

74.  Chapters  II.,  V.,  contain  little  more  about  the  trigonometric 
ratios  than  is  needed  in  the. solution  of  triangles.  In  this  and 
the  following  chapters  a  further  study  of  these  ratios  is  made. 
Although  the  results  of  this  study  are  not  applicable  to  such 
ordinary  practical  uses  as  the  measurement  of  triangles,  heights, 
and  distances,  yet  they  are  very  interesting  in  themselves,  and 
help  to  give  a  better  and  fuller  understanding  of  the  connection 
between  angles  and  trigonometric  ratios.  These  results  are  also 
useful  in  further  mathematical  work,  and  in  the  study  of  various 
branches  of  mechanical  and  physical  science.  In  reading  Chap- 
ters X.,  XI.,  acquaintance  will  be  made,  or  renewed,  with  some 
important  general  ideas  of  mathematics. 

75.  Function.  Trigonometric  functions.  If  a  number  is  so  re- 
lated to  one  or  more  other  numbers,  that  its  values  depend  upon 
their  values,  then  it  is  a  function  of  these  other  numbers.  Thus 
the  circumference  of  a  circle  is  a  function  of  its  radius ;  the  area 
of  a  rectangle  is  a  function  of  its  base  and  height ;  the  area  of  a 
triangle  is  a  function  of  its  three  sides. 

Note.  The  values  of  such  expressions  as  2  a;  —  5,  x'^  —  4  x  +  7,  logiox,  2*, 
depend  upon  the  values  given  to  x.  These  expressions  are,  accordingly, 
functions  of  x.  A  function  of  x  is  usually  denoted  by  one  of  the  symbols 
/(x),  F(x),  0(x),  etc.,  which  are  read  "the  /-function  of  x,"  "the  i^-func- 
tion  of  X,"  "the  Phi-innction  of  x,"  etc. 

The  trigonometric  ratios  of  an  angle  depend  upon  the  value 
(i.e.  magnitude)  of  the  angle.  On  this  account  the  trigonometric 
ratios  are  very  often  called  the  trigonometric  functions.  They  are 
also  frequently  called  the  circular  functions. 

The  trigonometric  (or  circular)  functions  include  not  only  the 

128 


74-76.]  ALGEBBAICAL  NOTE,  129 

six  functions  previously  discussed,  namely,  sme,  cosinej  tangent, 
cotangent,  secant,  cosecant,  but  also  three  others,  viz, : 

versed  sine  of  J.  =  1  —  cos  A,  written  vers  A, 
coversed  sine  of  ^  =  1  —  sin  A,  written  covers  A, 
suversed  sine  of  ^  =  1  +  cos  A,  written  suvers  A. 

The  versed  sine  is  used  not  unfrequently ;  the  latter  two  are 
rarely  used. 

EXAMPLES. 

Find  the  remaining  eight  trigonometric  functions  when : 
1.   sin  ^  =  .3.  2.   cos  J.  =  .4.  3.   tanj.  =  — 3.  4.   cot  ^  =  .7. 

5.   sec  J.  =  —  3.       6.   cosec  A  =  .8.       7.   vers  ^  =  1.5.         8.   vers  A  =  .5. 


9.   Show  that  ^^^^^  ^  -  ^f  ^'  ^  =  tan  A. 
1  —  vers  A 

10.   Show  that  cos  d  vers  ^  (1  +  sec  6)  =  sin^  6. 

76.  Algebraical  note. 

It  will  be  useful  to  have  an  idea  of  the  meaning  of  the  word  limit  as  used 
in  mathematics.    In  the  geometrical  series 

the  sum  of  2  terms  is  1|,  of  3  terms  is  If,  of  4  terms  is  1|,  of  5  terms  is  l|f . 
The  sum  of  the  series  varies  with  the  number  of  terms  taken  ;  and  the  greater 
the  number  of  terms  taken,  the  more  nearly  does  their  sum  approach  2.  It 
is  stated  in  arithmetic  and  algebra  that  the  sum  of  an  infinitely  great  num- 
ber of  terms  of  this  series  is  1  -4- (1  —  |),  i.e.  2.  This  simply  means  that,  by 
making  the  number  of  terms  as  great  as  one  please,  the  sum  can  be  made 
to  approach  as  nearly  as  one  please  to  2  ;  or,  in  other  words,  the  greater  the 
number  of  terras  taken,  the  more  nearly  does  their  sum  approach  the  value  2. 
This  idea  is  expressed  in  mathematics  in  slightly  different  language:  "The 
limit  of  the  sum  of  this  series  is  2."  In  geometry  (see  Note  C)  it  is  shown 
that  if  a  regular  polygon  be  inscribed  in  a  circle,  the  length  of  the  perimeter 
of  the  polygon  approaches  nearer  and  nearer  to  the  length  of  the  circle  as  the 
number  of  the  sides  of  the  polygon  is  increased  ;  also  the  area  of  the  polygon 
approaches  nearer  and  nearer  to  the  area  of  the  circle.  The  length  of  the 
circle  is  said  to  be  the  limit  of  the  length  of  the  perimeter  of  the  inscribed 
polygon,  and  the  area  of  the  circle  is  said  to  be  the  limit  of  the  area  of  the 
polygon,  as  the  number  of  its  sides  is  indefinitely  increased. 

Definition.     If  a  varying  quantity  approaches  nearer  and  nearer  to  a  fixed 
quantity  (or  given  constant)^  so  that  the  difference  between  the  two  quantl- 


130  PLANE  TlilGONOMETRY.  [Ch.  X. 

ties  may  become,  and  remain,  as  small  as  one  please,  then  the  fixed  quantity- 
is  called  the  limit  of  the  varying  quantity. 

The  following  algebraic  principles  are  required  in  some  of  the  articles  that 
follow :  I 

(a)   If  the  numerator  of  a  fraction  is  finite,  and  its  absolute  value  either 
remains  constant  or  increases  while  the  denominator  decreases,  then  the 

absolute  value  of  the  fraction  increases.     Thus,  if  in  ^,  a  either  remains 

a  ^ 

constant  or  increases  while  x  decreases,  then  -  increases  ;  e.g. 

_a  ^  _a_^   _a_^  ^^^^  values  10a,   100a,   1000a,    .... 

Tiy    T7(7     1000 

It  is  also  evident  that  the  smaller  x  becomes,  the  larger  does  -  become ; 

X 

and  that  -  approaches  an  exceedingly  great  value  when  x  approaches  zero. 

X 

In  other  words,  when  the  number  x  approaches  the  value  zero  as  its  limit,  then 
the  number  -  approaches  an  inconceivably  great  value  as  its  limit.      In 

X 

mathematics  numbers  of  the  latter  kind  are  each  denoted  by  the  word  infinity 
and  by  the  symbol  oo  ;  that  is,  the  symbol  co  denotes  any  number  which  is 
greater  than  any  number  that  can  be  assigned.  The  principle  just  stated, 
may  be  briefly  expressed  : 

If  x  =  0,  then  -  =  00, 

X 

in  which  the  symbol  =  indicates  the  following  reading  :  If  x  approaches  zero 
as  a  limit,  then  -  approaches  infinity  as  a  limit.  The  same  idea  is  also  ex- 
pressed thus : 


Limit  a 

a;  =  0x  =  °°' 

i.e.  the  limit  of  -,  when  x  is  zero,  is  infinity. 

X 

(b)  It  is  also  evident  that  as  x  increases,    -    decreases    (a   remaining 

^  a 

finite);  and  that  when  x  approaches  an  infinitely  great  value,  -  approaches 

zero.    That  is,  if 

x  =  cx>,  then  ^  =  0;  or,  ^^^^^^  =  0. 

(c)  If  X  approaches  zero,  then  -  approaches  zero,  so  long  as  a  does  not] 
approach  zero  ;  that  is,  ^ 

Limit  ^  _  rt 


77.] 


CHANGES  IN   TRIGONOMETRIC  FUNCTIONS. 


131 


77.  Changes  in  the  trigonometric  functions  as  the  angle  increases 
from  0°  to  360^  For  convenience  the  revolving  line  will  be  kept 
constant  in  length  in  the  following  explanations.  The  student 
should  try  to  deduce  the  changes  in  the  functions  for  himself, 
especially  after  reading  about  the  changes  in  the  sine. 

Change  in  sin  ^  as  ^  increases  from  0°  to  360°. 

If  OP  be  any  position  of  the  revolving  line,  then 


sin  XOP 


MP 

op' 


Now  OP  is  kept  the  same  in  length,  say  length  a,  as  XOP  in- 
creases from  0°  to  360°.  Hence,  in  order  to  trace  changes  in  the 
sine  as  the  angle  changes,  it  is  necessary  to  consider  only  the 
changes  in  MP.     Let  the  angle  be  denoted  by  A. 


When  ^  =  0,  OP  coincides  with  OB,  and  MP  =  0. 


0 


•.  sinO°  =  -  =  0. 
a 
As  OP  revolves  from  OX  to  O  Y,  MP  increases  in  length  and  is  positive. 

When  A  =  90°,  OP  coincides  with  0(7,  and  MP  =  a.      .'.  sin  90°  =  -  =  1. 

Hence,  as  the  angle  A  increases  from  0°  to  90°,  its  sine  increases  from 

0  to  1. 

As  OP  revolves  from  0  F  to  OXi,  MP  decreases  in  length  and  is  positive. 

When  A  =  180°,  OP  coincides  with  OBu  and  MP  =  0.    .-.  sin  180°  =  -  1=  0. 

Hence,  as  the  angle  A  increases  from  90°  to  180°,  its  sine  decreases  from 

1  to  0.  ^ 

As  OP  revolves  from  OXi  to  0  Yi,  MP  increases  in  length  and  is  negative  ; 
i.e.  MP  really  decreases. 


132  PLANE  TRIGONOMETRY.  [Ch.  X. 

When  A  =  270°,  OF  coincides  with  OCi,  MP  =  a  and  is  negative. 

.-.  sin  270-^  =  -^^  =  -1. 
a 

Hence,  as  the  angle  A  increases  from  180°  to  270°,  its  sine  decreases  from 

0  to  -  1. 

As  OP  revolves  from  O  Ti  to  OX,  MP  decreases  in  length  and  is  negative  ; 
i.e.  MP  really  increases. 

When  A  =  360°,  OP  coincides  with  OB,  and  MP  =  0.    .*.  sin  360°  =  -  =  0. 

Hence,  as  the  angle  A  increases  from  270°  to  360°,  its  sine  increases  from 
-  1  to  0. 

If  OP  continues  to  revolve,  then  the  sine  again  undergoes  the 
same  changes  in  the  same  order,  and  does  so  during  each  suc- 
cessive revolution. 

Change  in  cos^  as  A  increases  from  0°  to  360°, 

In   Fig.  67,  cos  XOP  =  -^—.      Hence,  in  order  to  trace  the 

changes  in  the  cosine  as  the  angle  changes,  it  is  necessary  to 
consider  only  the  changes  in  OMj  since  OP  is  kept  at  a  constant 
length  a. 

When  A  =  0°,  OP  coincides  with  OB,  and  OM  =  a.    .'.  cos  0°  =  -  =  1. 

As  OP  revolves  from  OX  to  0  F,  OM  decreases  in  length  and  is  positive. 

When  A  =  90°,  OP  coincides  with  OC,  and  OM  =0.    .-.  cos  90°  =  -  =  0. 

a 
Hence,  as  the  angle  A  increases  from  0°  to  90°,  its  cosine  decreases  from 

1  too. 

As  OP  revolves  from  OF  to  OXi,  Oilf  increases  in  length  and  is  negative, 
i.e.  Oi¥^ really  decreases. 

When  A  =  180°,  OP  coincides  with  0-Bi,  OM  =  a,  and  is  negative. 

.-.  cos  180°==^  =  -!. 
a 

Hence,  as  the  angle  increases  from  90°  to  180°,  its  cosine  decreases  from 
0  to  -  1. 

On  proceeding  in  the  same  manner,  the  student  will  discover  that : 
As  A  increases  from  180°  to  270°,  cos  A  increases  from  —  1  to  0 ; 
As  A  increases  from  270°  to  360°,  cos  A  increases  from  0  to  1. 

If  OP  continues  to  revolve,  then  the  cosine  again  undergoes  the 
same  changes  in  the  sf'^ne  order,  and  does  so"  during  each  succes- 
sive revolutioT" 


77.]  CHANGE  IN  TAN  A,  133 

Change  in  tan  JL  as  A  increases  from  0°  to  360°. 

MP 

In  rig.  67,  tan  XOP  = .      Hence,  in  order  to  trace   the 

changes  in  the  tangent  as  the  angle  changes,  it  is  necessary  to 
consider  the  changes  in  MP  and  OM. 

When  A  =  0°,  OP  coincides  with  OB,  MP=0,  OM=a.    .-.  tan  ^  =  -=  0. 

As  OP  revolves  from  OX  to  0  F,  MP  increases  and  OM  decreases,  and 
both  are  positive  ;  hence,  tan  A  increases. 

When  A  =  90°,  OP  coincides  with  00,  MP  =  a,  OM=:0. 


As  OP  revolves  from  OF  to  OXi,  MP  decreases  and  is  positive,  OM  in- 
creases in  length  and  is  negative  ;  hence,  tan  A  decreases  in  magnitude  and 
is  negative  ;  i.e.  tan  A  really  increases.  [When  OP  passes  at  OF  from  the 
first  quadrant  into  the  second,  the  value  of  the  tangent  changes  from  +  co  to 
—  00,  for  Oilf  changes  its  sign  from  +  to  — .] 

When  A  =  180°,  OP  coincides  with  07?i,  MP=0,  0M  =  -  a. 

.-.  tanl80°=  — =  0. 
—  a 

Hence,  as  A  increases  from  90°  to  180°  tan  A  increases  from  —  oo  to  0. 
On  proceeding  in  the  same  manner  the  student  will  discover  that : 
As  A  increases  from  180°  to  270°,  tan  A  increases  from  0  to  +  co ; 
As  A  increases  from  270°  to  360°,  tan  A  increases  from  —  oo  to  0. 

If  OP  continues  to  revolve,  then  the  tangent  again  undergoes 
the  same  changes  in  the  same  order,  and  does  so  during  each 
successive  revolution. 

In  the  same  way  as  above,  the  student  can  trace  the  changes  in 
sec  A,  cosec  A,  cot  A,  as  A  increases  from  0°  to  360°.  The  changes 
in  these  functions  can  also  be  deduced  from  the  results  obtained 
for  sin  A,  cos  A,  tan  A,  and  the  relations 

sec  A  = 7,  cosec  A  =  -: — -,  cot  A  = 


cos  A  sin  A  tan  A 

The  results  are  collected  in  the  following  table :  * 

*This  method  of  indicating  the  changes  in  the  trigonometric  functions 
i ;  that  given  in  Loney's  Plane  Trigonometry,  p.  67. 


134 


PLANE  TRIGONOMETRY. 


[Ch.  X. 


In  the  second  quadrant  the 
sine           decreases  from       1  to 
cosine        decreases  from       0  to 

r 

In  the  first  quadrant  the 
0    sine           increases  from 
-1     cosine        decreases  from 

0  to 

1  to 

1 

0 

tangent     increases  from  - 
cotangent  decreases  from 
secant       increases  from  - 

-Qo  to 
0  to 

-QO   to 

0 

-00 

-1 

tangent      increases  from 
cotangent  decreases  from 
secant        increases  from 

0  to 

00  to 

1  to 

00 

0 

00 

cosecant    increases  from 

1  to 

00 

cosecant    decreases  from 

00  to 

1 

v 

In  the  third  quadrant  the 
sine           decreases  from 

0  to 

-1 

0 

In  the  fourth  quadrant  the 
sine            increases  from    — !  to 

0 

cosine        increases  from    - 

-1  to 

0 

cosine        increases  from 

0  to 

1 

tangent     increases  from 
cotangent  decreases  from 
secant       decreases  from   - 

0  to 

00  to 

-1  to 

00 

0 

—  00 

tangent      increases  from 
cotangent  decreases  from 
secant        decreases  from 

—  00  to 
0  to 
ooto 

0 

-00 

1 

coseoaiUt    increases  from    - 

-co  to 

-1 

cosecant    decreases  from 

-1  to 

—  OQ 

Note.  It  should  be  observed  that  the  algebraic  sign  of  each  function 
changes  when  the  function  passes  through  either  of  the  values  zero  and 
infinity. 

Ex.  Trace  the  changes  in  the  versed  sine  as  the  angle  changes  from  0°  to 
360°. 

78.  Periodicity  of  the  trigonometric  functions.  It  has  been  seen 
in  Arts.  40-44  that  all  angles  coterminal  with  XOP  have  the  same 
ratios.  That  is,  the  same  ratios  as  XOP  has, 
are  obtained  each  time  that  the  revolving  line 
returns  to  the  position  OP,  no  matter  how  many 
complete  revolutions  in  the  positive  or  negative 
X^      0  X     direction  it  may  make  in  the  meantime.     In  the 

Fio.  68.  last  article  it  was  pointed  out  that  the  sine,  for 

instance,  always  goes  through  all  its  changes  (the 
cycle  of  changes,  namely,  0  to  1,  1  to  0,  0  to  —  1,  —  1  to  0)  in 
the  same  order  when  the  turning  line  revolves  from  the  position 
OX  through  the  angle  360°  or  2  tt.  According  to  the  opening 
remarks  of  this  article,  all  angles  which  differ  by  any  integral  multi- 
ple (positive  or  negative)  o/  2  tt  radians  have  the  same  sine.  These 
facts  are  expressed  mathematically  by  saying : 

The  sine  is  a  periodic  function,  and  the  period  of  the  sine  is  2  ir. 


78,79.]  PERIODICITY  OF  FUNCTIONS,  135 

Similar  considerations  show  that  the  cosine,  secant,  cosecant,  are 
periodic  functions,  and  that  each  of  them  has  a  period  2  tt. 

The  tangent  and  cotangent,  however  (Art.  77),  go  through  all 
their  changes,  while  the  angle  increases  by  180°  or  ir  radians. 
Hence,  the  period  of  the  tangent  and  cotangent  is  tt. 

Note  1.  These  properties  may  be  expressed  as  follows,  w  denoting  any- 
positive  or  negative  whole  number,  and  x  being  any  angle : 

sin  X  =  sin(2  imr  +  x),   cos  x  =  cos(2  mir  +  x), 
and  similar  for  sec  x,  cosec  x  ; 

tan  X  =  tan(m7r  +  x),     cot  x  =  cot(?7i7r  +  x). 

Note  2.  (Algebraic.)  When  a  function  /(x)  has  the  property  that 
f{x)  =  f(x  +  k),  in  which  x  can  have  any  value  and  A;  is  a  constant,  the 
function  /(x)  is  said  to  be  a  periodic  function.  If  k  is  the  least  quantity  for 
which  this  equation  is  true,  then  k  is  called  the  period  of  the  function. 

If  f{x)  =  f{x  +  k),  then  f(x)  =  f(x  +  nk),  n  being  any  positive  or  nega- 
tive whole  number.  For  f(x  +  k)=f(x  +  k  +  k)=f{x  +  2  A;),  and  so  on. 
Also,  since  f(x)=f(x  +  k)  for  all  values  of  x,  this  equation  holds  when 
X  — A;  is  put  for  x;  that  is,  f(x—k)  =  f{x).  Similarly,  /(x-2A:)  = 
f{x  —  k)=f(x)^  and  so  on. 

Note  3.  It  has  been  shown  above  that  each  of  the  trigonometric  func- 
tions has  but  one  period,  namely,  tt,  in  the  case  of  the  tangent  and  cotangent, 
and  2  TT  in  the  case  of  each  of  the  other  functions.  Hence,  the  trigonometric 
functions  are  singly  periodic  functions.  Functions  which  have  more  than 
one  period  appear  in  some  branches  of  higher  mathematics.  For  instance, 
certain  functions  called  elliptic  functions  have  two  periods,  and,  accordingly, 
are  said  to  be  doubly  periodic.       See  Questions  on  Chap.  X.,  Ex.  16. 

79.  The  old  or  line  definitions  of  the  trigonometric  functions.    The 

trigonometric  functions  were  formerly  considered  as  belonging  to 

arcs   rather   than   to  angles,  and  were 

certain  lines  related  to  these  arcs.     Let 

APB  be   a  circle   described  with   any 

radius  R  aboufc  0  as  a  centre.    Let  OA, 

OB,  be  at  right  angles  to  each  other, 

and  let  AP  be  any  arc  having  A  for  its 

initial  point.     Draw  OP;  from  P  draw 

PM  at  right  angles  to  OA ;  through  A 

draw  a  tangent  AT  to  meet   OP  pro-  Em.  69. 

duced  in  T;  through  B  draw  a  tangent 

C7\  to  meet  OP  produced  in  T^ ;  from  P  draw  PM^  at  right  angles 


136  PLANE  TRIGONOMETRY.  [Ch.  X. 

to  OB.  The  lines  MP,  AT,  OT,  AM,  were  called  respectively,  the 
sine,  tangerit,  secant,  cover sed  sine,  of  the  arc  AP;  and  M^P,  BT^, 
OTi  (the  sine,  tangent,  secant,  of  the  complementary  arc  PB) 
were  called  respectively,  the  cosine,  cotangent,  cosecant,  of  the  arc 
AP.     These  definitions  are  expressed  in  words  as  follows : 

The  sine  of  an  arc  is  a  straight  line  drawn  from  one  extremity 
of  the  arc  perpendicular  to  the  radius  passing  through  the  other 
extremity.  The  tangent  of  an  arc  is  a  straight  line  touching  the 
arc  at  one  extremity,  and  limited  by  the  radius  produced  through 
the  other  extremity.  The  secant  of  cm  arc  is  thG  straight  line 
joining  the  centre  of  the  circle,  and  the  further  extremity  of  the 
tangent  drawn  at  the  origin  of  the  arc. 

The  sine,  tangent,  and  secant  of  the  complement  of  an  arc  are 
called  the  cosine,  cotangent,  and  cosecant  of  that  arc. 

Since  the  arc  measures  the  angle  at  the  centre  (the  number  of 
degrees  in  this  arc  is  the  same  as  the  number  of  degrees  in  the 
subtended  angle),  these  lines  were  also  called  the  sine,  cosine, 
•  ••,  of  the  central  angle  AOP  measured  by  the  arc  AP.  These 
lines  were  known  as  the  trigonometric  lines. 

Note.  By  "  the  length  of  a  line  "  is  meant  the  number  of  units  of  length 
which  it  contains.  The  lengths  of  these  lines  depend  on  the  length  of  the 
radius  of  the  circle.,  as  well  as  on  the  magnitude  of  the  central  angle  subtended 
by  the  arc.  Hence  it  was  necessary  to  specify  the  radius  when  the  functions 
were  discussed.  This  inconvenience  has  led  to  the  adoption  of  the  ratio 
definitions. 

In  Fig.  69  let  R  denote  the  length  of  the  radius.  Then,  on 
using  the  ratio  definitions, 

sin  ^0P  =  ^,   tan  ^0P  =  —,   sec  ^0P  =  ^. 
R  R  R 

Hence,  the  ratio  definitions  of  the  trigonometric  functions  can 
be  derived  from  the  line  definitions  by  dividing  the  lengths  of  the 
lines  by  the  length  of  the  radius.  Jf  the  length  of  the  radius  is 
unity,  then  the  lengths  of  the  lines  used  m  the  line  definitions  are 
equal  to  the  ratios  in  the  ratio  definitions.  This  suggests  a  geo- 
metrical or  graphical  method  of  representing  the  trigonometric 
functions,  which  is  shown  in  the  next  article. 


80.]  GEOMETRICAL   BEPBESENTATION.  137 

80.   Geometrical   representation    ef   the   trigonometric  functions. 

Let  a  circle  of  radius  equal  to  unity  be  drawn.     This  circle  is 

called  a  unit-circle.      Let  the  construction 

described  in  Art.  79  be  made  for  each  of 

the  angles  AOP^,  AOP^,  AOP^,  •••.     In  any 

circle  the  lines  M^P^,  M^P^,  M^P^,   "•,  are 

proportional    to,    and    hence    represent    the 

sines  of  these  angles, 

the  lines  AT^,  ATo_,  AT^,  "-, 

represent  the  tangents  of  these  angles, 

the  lines  0J\,  OT^,  OT^,  .••, 

represent  the  secants  of  these  angles,  ^^^'  '^^^ 

and  so  on  for  the  other  ratios.  In  the  unit-circle,  however,  the 
measures  of  these  lines,  the  radius  being  the  unit  of  length,  are  the 
very  same  numbers  as  the  respective  ratios  mentioned.  In  the  unit- 
eircle  also,  the  linear  measure  of  the  arc  is  the  same  as  the  radian 
measure  of  the  angle  which  it  subtends.     [See  Art.  73,  Note  3.] 

Suggested  Exercises.  (1)  By  means  of  the  lines  on  the  unit-circle, 
trace  the  changes  in  the  trigonometric  functions  as  the  angle  changes  from 
0°  to  90°.     Compare  the  results  with  those  of  Art.  77. 

(2)  For  particular  values  of  the  angle  AOPi,  measure  the  lengths  of  the 
related  lines  on  the  unit-circle,  and  compare  the  results  with  the  values  given 
in  the  tables  of  natural  sines  and  tangents. 

Note  1.  The  origin  of  the  terms  circular  functions,  tangent.,  secant^  is 
apparent  from  Art.  79. 

Note  2.  The  name  sine  comes  from  the  Latin  word  sinus,  which  was 
the  translation  of  the  Arabic  word  for  this  trigonometric  function.  The 
Arabic  word  for  the  sine  resembled  a  word  meaning  an  indentation  or  gulf. 

Note  3.  In  trigonometry  the  Greeks  used  the  ivhole  chord  FiQ  instead 
of  the  /iaZ/-chord  or  sine.  For  example,  Ptolemy,  the  celebrated  astronomer 
who  flourished  about  125-151  a. d.,  gave  a  table  of  chords  in  Book  L  of  the 
Almagest,  his  work  on  astronomy.  The  Hindoos,  on  the  other  hand,  always 
used  the  half-chord  or  sine.  The  Arabian  astronomer,  Al  Battani  (or  Alba- 
tagnius)  (877-929),  in  his  work  The  Science  of  the  Stars,  like  the  Hindoos 
determined  angles  "by  the  semi-chord  of  twice  the  angle,"  i.e.  by  the  sine 
of  the  angle,  taking  the  radius  as  unity.  The  translation  of  this  work  into 
Latin  in  the  twelfth  century  introduced  the  word  sine  into  trigonometry. 
The  Hindoo  sine  was  finally  adopted  in  Europe  in  preference  to  the  Greek 
chord  in  the  fifteenth  century.     [See  Art.  12,  foot-note.] 


138 


PLANE  TBIGONOMETRY. 


[Ch.  X. 


81.    Graphical  representation  of  functions. 

Graphical  representation.  The  different  values  which  a  varying  quan- 
tity takes,  are  often  represented  by  means  of  a  curve.  Many  illustrations 
can  he  given  of  the  graphical  representation  of  various  things  whose  values 
can  be  denoted  by  means  of  numbers.  For  example,  the  curve  in  Fig.  71 
shows  the  record  of  the  barometer  at  Ithaca  from  May  22  to  May  29,  1899. 


29.7 

" 

" 

~ 

" 

"" 

29.5 
29.1 
29.3 
29.2 
29.1 
29.0 
28.9 

>■ 

> 

,^^ 

- 

«» 

_ 

-' 

S 

*" 

s 

> 

*«^ 

^^> 

H 

EIGHT  OF  Bt 

ROV.ETER 

> 

^ 

""X^ 

AT  ITHACA,  N.Y. 

MAY  22  TO  MAY  29 

1899 

- 

-r 

28.7 
2a6 

llllllllllll 

L 

Inches 
29.7 
29.6 
29.5 
29.4 
29.3 
29.2 
29.1 
29.0 


12  3C9123C9123C9123  0  9123C9123C9123C9123C9123C9123C9123C91236  9123C9123C912 

M.         P.M.       Al.        P.M.       M.         P.M.       M.        P.M.        M.       P.M.        M.        P.M.        M.        P.M.        M. 

Monday         Tusiday  Wednesday  Thursday  Friday  Saturday  Sunday        Monday 

May  22  May  23  May  2i  May  25  May  26  May  21  May  2S        May  29 

Fig.  71. 

In  this  figure  an  hour  is  represented  by  a  certain  length,  and  the  lengths  rep- 
resenting hours  are  measured  along  a  horizontal  line.  Each  inch  of  height 
of  the  barometer  is  also  represented  by  a  certain  length.  At  the  points  cor- 
responding to  the  successive  times  perpendiculars  are  drawn-,  the  lengths  of 
which  represent  the  heights  of  the  barometer  at  the  respective  times.  (In  the 
figure  the  position  of  the  horizontal  line  marked  29,  represents  the  upper  ends 
of  heights  of  29  inches.)  The  smooth  curve  drawn  through  the  extremities  of 
the  perpendiculars  is  the  barometric  curve  or  curve  of  barometric  heights  for 
the  period  May  22  to  May  29,  1899.  This  curv^  will  give  to  most  persons  a 
clearer  and  more  vivid  idea  of  the  range  and  variation  of  the  height  of  the 
barometer  during  this  period  than  a  column  of  numbers  of  inches  of  heights 
is  likely  to  give.  If  the  scales  used  in  representing  the  hours  and  the  heights 
of  the  barometer  were  changed,  then  the  curve  would  be  somewhat  altered, 
but  its  general  appearance  would  remain  the  same. 

The  graph  of  a  function.     The  graph  of  a  function  of  as,  say  /(a;),  is 

obtained  in  the  following  way:  Take  a  horizontal  line  XiOX;  choose  a 
point  0,  from  which,  distances  representing  the  different  values  of  x  are 
measured  along  the  line  ;  measure  positive  values  of  x  toward  the  right 
from  0,  and  negative  values  toward  the  left.  At  particular  points  of  XiOX, 
at  convenient  distances  apart,  draw  perpendiculars  to  represent  the  values 
of  f(x)  at  the  respective  points.  Draw  the  perpendiculars  upward  from 
XiOX  when  the  values  of  f(x)  are  positive,  and  downv/ard  when  these 
values  are  negative„  The  smooth  curve  drawn  through  the  extremities  of 
these  perpendiculars  is  the  graph  of  /(x).     The  nearer  the  perpendiculars 


Sl-82.]       GRAPHS  OF  TRtGONOMETHlC  FUNCTIONS. 


139 


are  to  one  another,  the  better  is  the  graph.  For  example,  the  function  2  x 
is  represented  (for  certain  values  of  x)  by  Fig.  72,  the  function  ^x^,  by 
Fig.  73,  the  function  Vx,  by  Fig.  74.  The  pupil  is  advised  to  construct  these 
graphs  by  following  the  method  just  described  above. 


Graph  of  V^ 
Fig.  74. 


Exs.  Draw  the  graphs  for  3  oc,  ^  jc,  4  a;  +  5,  4  a;  —  5,  ^  x^,  i  x^.     v25  —  x'^. 

Note.  The  notion  of  representing  a  function  by  a  curve  is  the  funda- 
mental notion  in  algebraic  geometry,  or,  as  it  is  usually  termed,  analytic 
geometry.  This  geometry  was  invented,  in  the  form  in  which  it  is  now 
known,  by  the  philosopher  and  mathematician,  Ren6  Descartes  (1596-1650), 
and  first  published  by  him  in  1637.  This  article  may  be  regarded  as  a  short 
lesson  in  the  subject. 

82.   Graphs  of  the  trigonometric  functions. 

Graph  of  sin  0.  In  order  to  draw  the  graph  of  sin  9  take  dis- 
tances, measured  from  0  along  the  line  X^OX,  to  represent  the 
number  of  radians  in  the  angle  0.  At  points  (not  too  far  apart) 
on  X^OX  draw  perpendiculars  to  represent  the  sines  of  the  angles 
corresponding  to  these  points.  The  smooth  curve  drawn  through 
the  extremities  of  these  perpendiculars  will  be  the  graph  of  the 
sine.  Thus,  for  example,  let  a  radian  be  represented  by  a  unit 
length,  and  let  the  ratio  unity  be  also  represented  by  a  unit  length. 
Then  (see  Fig.  75)  angle  tt  {i.e.  180°)  is  represented  by  OM^  =  3|. 
The  perpendiculars  at  0  and  M^  are  zero,  since  sin  0  =  0  and 
sin7r  =  0.  Erect  perpendiculars  equal  to  •••,  sin  30°,  sin  45°, 
sin  60°,  sin  90°,  •••,  for  instance,  at  the  points  corresponding  to  •••, 


7r     TT     TT     TT 

G'  4'  3'  2' 


.,  (i.e.  .-.,  30°,  45°,  60°,  90°,  ••.,)  respectively.     Do  the 


140  PLANE  TRIGONOMETRY,  [Ch.  X. 

same  at  points  between  L  and  M^,  and  draw  the  smooth  curve 
OQ\Mi  through  the  extremities  of  the  perpendiculars.  The  suc- 
cessive perpendiculars  from  tt  to  2  tt  are  the  same  in  length  as 
those  from  0  to  tt,  but  negative.  From  2  tt  to  4  tt  the  values  of 
the  sine  are  repeated  in  the  same  order  as  from  0  to  2  tt.  Hence, 
the  graph  of  the  sine  can  be  obtained  by  merely  successively 
reproducing  the  double  undulation  OG1M1G2M2,  as  indicated  in 
Fig.  75.     This  is  called  the  curve  of  sines,  sine  curve,  or  sinusoid. 

Note  1.  The  unit  circle  (Art.  80)  will  be  of  service  in  drawing  the 
graphs  of  the  sine  and  the  other  trigonometric  functions.  For,  if  the  scales 
for  radians  and  ratios  be  those  adopted  above,  then  the  horizontal  distances 
from  0  will  be  equal  to  the  lengths  of  the  arcs  (Art.  73,  Note  2),  and  the 
lengths  of  the  perpendiculars  will  be  the  lengths  of  the  lines  in  the  line  defini- 
tions (Art.  79). 

Note  2.  If  tt  radians  {i.e.  180°)  be  represented  by  a  length  different  from 
that  adopted  in  Fig.  75,  then  the  graph  of  the  sine  will  differ  somewhat  from 
Fig.  75,  but  its  main  features  will  be  the  same  as  in  that  figure.  Figures  76 
and  77  show  portions  of  the  graph  of  sin  6  when  tt  is  represented  on  two  other 

scales,  while  the  sin-  {i.e.  1)  is  represented  by  a  unit  length.     Hence  the 

curve  of  sines,  or  the  sinusoid,  may  be  defined  as  the  curve  in  which  hori- 
zontal distances  measured  on  a  certain  line  are  proportional  to  an  angle,  and 
the  perpendiculars  to  this  line  are  proportional  to  its  sine. 

Ex.     Draw  the  graphs  for  cos  6,  tan  6,  cot  0,  sec  6,  cosec  6. 

Graph  of  cos  0.  On  using  the  same  scales  for  radians  and  ratios 
as  those  adopted  in  Fig.  75,  the  graph  of  cos  6  takes  the  form 
shown  in  Fig.  78.  It  is  the  same  as  the  graph  of  sin  0  in  Fig.  75 
would  be  if  0  and  the  other  points  in  XiOX  were  all  moved  a 
distance  1  tt  toward  the  right.  This  might  have  been  expected, 
since  the  sine  of  an  angle  is  equal  to  the  cosine  of  its  comple- 
ment. The  values  of  the  sine  and  the  cosine  alike  range  from 
+  1  to  -  1. 

Graph  of  tan  0.  On  using  the  same  scales  for  radians  and  ratios 
as  have  been  adopted  in  Fig.  75,  the  graph  of  tan  0  takes  the 
form  shown  in  Fig.  79. 

Graph  of  sec  0.  On  using  the  same  scales  for  radians  and  ratios 
as  have  been  adopted  in  Fig.  75^  the  graph  of  sec  0  takes  the 
form  shown  in  Fig.  80. 


62.] 


GRAPHS   OF  SIN  6  AND  COS  6. 


141 


\ 

s 

/ 

CO 

f 

^1- 

\ 

\ 

8    00* 


142 


I*LANE  TRIGONOMETRY, 


[Ch.  X. 


ft 

I 

i 

1 

1 

1 

/ 

/j 

2 

/  1 

/ 

y 

! 

37r 

^^ 

1 

1/ 

^OJL 

77 

A 

37r 

/ 

/- 

|57r 

/ 

X 

Graph  of  tan  0 
Fig.  79. 


Xy^  ^.E. 


Graph  of  sea  d 
Fig.  80u 


83.] 


RELATIONS  BETWEEN  6,   SIN  6,    TAN  6.  143 


EXAMPLES. 

1.  Draw  the  graph  for  cot  d,  using  the  scales  adopted  in  Fig.  75. 

2.  Draw  the  graph  for  cosec  6,  using  the  scales  adopted  in  Fig.  75. 

3.  Construct  various  graphs  for  sin  0,  cos  d,  tan  0,  cot  d,  sec  6,  cosec  d,  by 
varying  the  scales  used  in  representing  radians  and  ratios. 

4.  Draw  graphs  for  : 

(a)  sin  X  +  cos  x,    (b)  sin  x  —  cos  x,    (c)  sin  2  x,    (d)  cos  2  x. 

83.  Relations  between  the  radian  measure,  the  sine,  and  the  tan- 
gent of  an  acute  angle. 

A.  If  e  be  the  radian  measure  of  an  acute  angle,  then  sin  0 <e <tan 0. 

Let  angle  AOP=0,  make  the  angle 
AOR  equal  to  0,  and  with  any  radius 
r  describe  the  arc  QBE  about  0  as  a 
centre.  Draw  the  chord  QR  inter- 
secting OB  in  Mj  and  draw  the  tan- 
gents at  Q  and  M.  By  geometry, 
arc  QB  =  arc  BR,  QR  is  at  right 
angles  to  OB,  MQ  =  MR,  the  tan- 
gents at  Q  and  R  intersect  at  a  point 
Ton  OA,  QT=RT. 

Note.  If  r  =  1,  that  is,  if  QB  is  an  arc 
of  a  unit  circle,  then  the  linear  measures  of 
MQ,  BQ,  TQ  are  equal  to  sin  6,  d,  tan  6, 
respectively. 

It  can  be  easily  shown  by  mechanical  means  that 

MQ<BQ<TQ.  (1) 

For  suppose  that  pegs  are  placed  at  Q,  B,  jT,  and  that  a  string  is  drawn 
taut  from  ^  to  i? ;  suppose  that  another  string  is  drawn  from  Q  to  B,  but 
constrained  to  lie  on  the  circular  arc  QB,  like  a  string  stretched  along  the  tire 
of  a  wheel.  Also  let  a  third  string  be  drawn  taut  from  j^  to  Q,  but  passed 
over  the  peg  at  T.  Then  it  is  obvious  that  the  first  of  the  three  strings  is  the 
shortest,  and  the  third  is  the  longest.  The  second  string  cannot  be  drawn  away 
from  the  arc  QBB  without  being  stretched,  and  if  peg  T  were  removed,  the 
string  Q  TB  would  lie  loosely  on  the  arc  QBB.  Since  QMB  <  QBB  <  QTB, 
then  MQ  <^BQ<_TQ;  that  is,  r  sin  ^  <  r^  <  r  tan  6.    Hence,  sin  ^  <  ^  <  tan  6. 

The  truth  of  A  may  be  perceived  from  the  following  mathe- 
matical consideration.     Draw  the  chord  QB.     Evidently, 

area  triangle  OQB  <  area  sector  OQB  <  area  triangle  OQT. 


ym.  81. 


144  PLANE  TBIGONOMETRY.  [Ch.  X. 

That  is,         iOB'MQ<^OQ'  arc  BQ  <  -J-  OQ-QT-, 
or,  ^r  '  rsinO  <^r  '  r9<^r  'V  tan  6. 

Hence,  sin  0  <  0  <  tan  0.  (1) 

B.   When  angle  0  approaches  zero,  each  of  the  ratios  ^^,  tan0^ 

0         0 
approaches  unity  as  a  limit.     On  dividing  each  of  the  members  of 

the  inequality  (1)  by  sin  0,  there  is  obtained 


sin  6     cos  9 

But  when  0  approaches  zero,  cos  6  approaches  unity  a^s  a  limit 
(^Art.  77).  Hence,  when  6  approaches  zero,  -; —  must  also  ap- 
proach unity  as  a  limit;   that  is,  the  limit  of  ^IIL-  is  1. 

0 
On  dividing  the  members  of  (1)  by  tan  9,  there  is  obtained 

cos6l<-^<l. 
tan^ 

As  before,  when  9  approaches  zero,  cos  0  approaches  unity  as  a 
9 
limit,  and  hence approaches  unity  as  a  limit ;  i.e.  the  limit 

tan  9  .  *^^ 

of  — -—  is  1.     These  results  may  be  briefly  expressed : 

Limit/sin0\  _  ^  .    Limit /tan  0\     -g  zox 

These  are  two  of  the  most  important  theorems  in  elementary 
trigonometry;  they  are  frequently  employed  both  in  practical 
work  and  in  pure  mathematics. 

A  very  important  corollary  to  (2)  is  the  following : 
If  9  be  the  radian  measure  of  a  very  small  angle,  then  9  can  be 
used  for  sin  9  and  tan  9  in  calculations. 

For  instance,  sin  10"  to  12  places  of  decimals  is  .000048481308.  This  is 
also  the  radian  measure  of  10"  to  12  places  of  decimals.  The  radian  meas- 
ures, sines,  and  tangents,  of  angles  from  0°  to  6°,  agree  in  the  first  three  places 
of  decimals.     For 

radian  measure  6°  =  (.10472)  =  .105 ;  sin  6°  =  (.10453)  =  .105; 
tan  6°  =  (.10510)  =  .105. 


I 


pi" 

83.] 


EXAMPLES,  145 


EXAMPLES. 

1.  Find  the  angle  subtended  by  a  man  6  ft.  high  at  a  distance  of  half  a 
mile. 

Here,  ^  =  tan  ^  =  ——  =  — .  0. . 

2640      440  — ^^  "l«^^ 

2640  Ft. 

Now        il  =  J-  X  1^°  =  7  X  180°  ^  ^,  4g„  g^  Fia.,  83. 

440      440        IT        22  X  440 

2.  What  must  be  the  height  of  a  tower,  in  order  that  it  subtend  an  angle 
1°  at  a  distance  of  4000  ft.  ? 

-^  =  tan  1°  =  radian  measure  1"  =  .^^  =  — — —  —       

4000  180      7  X  180  4000 

.^^22x4000^gg3^^^^  FIG.  83. 

7  X  180 

3.  Verify  the  statements  made  in  Art.  11,  Note  1.    (Take  tt  =  3. 14159265.) 

4.  The  moon's  mean  angular  diameter  as  observed  at  the  earth  is  31'  5", 
and  its  actual  diameter  is  about  2160  miles.  Pind  the  mean  distance  of  the 
moon.     How  many  full  moons  would  make  a  chaplet  across  the  sky  ? 

5.  Taking  the  earth's  equatorial  radius  as  3963  mi.,  find  the  angular 
semi-diameter  of  the  earth  as  it  would  appear  if  observed  from  the  moon. 
Compare  the  relative  apparent  sizes  of  the  moon  as  seen  from  the  earth,  and 
the  earth  as  seen  from  the  moon. 

6.  The  semi-diameter  of  the  earth  as  seen  from  the  sun  is  very  nearly 
8". 8.  (See  Art.  11,  Note  1.")  What  is  the  sun's  distance  from  the  earth, 
the  radius  of  the  earth  being  assumed  as  4000  miles  ? 

7.  At  least  how  many  times  farther  away  than  the  sun  is  the  nearest 
fixed  star  a  Centauri,  at  which  the  mean  distance  between  the  earth  and  sun 
(about  92,897,000  miles)  subtends  an  angle  something  less  than  1"  ?  How 
long,  at  least,  will  it  take  light  to  come  from  this  star  to  the  earth  ? 

8.  Find  approximately  the  distance  at  which  a  coin  an  inch  in  diameter 
must  be  placed  so  as  just  to  hide  the  moon,  the  latter's  angular  diameter 
being  taken  31'  5". 

9.  The  inclination  of  a  railway  to  a  horizontal  plane  is  50'.  Find  how 
many  feet  it  rises  in  a  mile. 

10.  Find  the  angle  subtended  by  a  circular  target  4  ft.  in  diameter  at  a 
distance  of  1000  yd. 

11.  Find  the  height  of  an  object  whose  angle  of  elevation  at  a  distance  of 
900  yd.  is  1°. 

12.  Find  the  angle  subtended  by  a  pole  20  ft.  high  at  a  distance  of  a  mile. 

13.  Exs.  5,  6,  Art.  34  b. 

N.B.     Questions  and  exercises  suitable  for  practice  and  review  on  the 
subject-matter  of  Chapter  X.  will  be  found  at  page  195. 


CHAPTER    XI. 

GENERAL    VALUES.      INVERSE    TRIGONOMETRIC 
FUNCTIONS.      TRIGONOMETRIC    EQUATIONS. 

84.  General  values.  Articles  40-43  should  be  reviewed  care- 
fully before  this  chapter  is  taken  up.  It  has  been  seen  in  these 
articles  that  all  co-terminal  angles  have  the  same  trigonometric 
ratios.  It  was  also  pointed  out  in  Art.  43  that  two  sets  of  co-ter- 
minal angles,  each  set  being  infinite  in  number,  correspond  to 
any  given  ratio.  For  example,  in  Art.  42,  Ex.  1,  Fig.  38,  any 
angle  whose  terminal  line  is  either  OP  or  OPi  has  a  sine,  J ;  in 
Ex.  2,  Fig.  39,  any  angle  whose  terminal  line  is  either  OP  or  OP^ 
has  a  tangent,  —  |.  One  of  the  objects  of  this  chapter  is  to 
derive  expressions  or  formulas  that  will  include  all  angles  which 
have  the  same  sine,  cosine,  tangent,  cotangent,  secant,  cosecant, 
respectively.  These  general  expressions  are  sometimes  called 
general  values.  The  student  is  advised  to  deduce,  after  reading 
Art.  85,  the  general  values  for  cosine,  tangent,  etc.,  without  the 
help  of  the  book. 

85.  General  expression  for  all  angles  which  have  the  same  sine. 
Let  s  be  the  given  value  of  the  sine.  It  is  required  to  find  an 
expression  that  will  represent  and  include  every  angle  whose 
sine  is  s.     All  the  angles  whose  sines  are  equal  to  s  can  be  repre- 


sented  geometrically,  as   shown   in   Art.  42,   and   indicated   in 
Figs.  84,  85.     In  Fig.  84,  s  is  positive ;  in  Fig.  85,  .s  is  negative. 
Let  XOP  be  the  least  positive  angle  whose  sine  is  s.     Let 

146 


84,  85.]  GENERAL    VALUES.  147 

XOP=A ;  then  XOPi  =  180°  —  A.  Every  angle  whose  terminal 
line  is  either  OP  or  OPy  has  its  sine  equal  to  s.  Now  all  angles 
having  OP  for  a  terminal  line  are  obtained  by  adding  all  num- 
bers of  complete  revolutions  (positive  and  negative)  to  XOP. 
Hence,  these  angles  are  represented  by 

m  .  360°  +  A,  i.e.  2  m  •  180°  +  A,  (1) 

in  which  m  is  any  positive  or  negative  whole  number. 

Similarly,  all  angles  having  OPi  for  a  terminal  line  are  repre- 
sented by 

m  .  360°  -1-  (180°  -  A),  i.e.  (2  m  +  1)  180°  -  A.  (2) 

An  expression  that  will  include  both  sets  of  angles,  (1)  and  (2), 
'will  now  be  obtained.  In  the  expression  (1),  the  coefficient  of 
180°  is  even,  and  the  sign  of  A  is  positive;  in  (2),  the  coefficient 
of  180°  is  odd,  and  the  sign  of  A  is  negative.  Hence,  n  being  any 
positive  or  negative  whole  number,  the  expression 

wl80°  +  (-ir^,  (3) 

includes  the  angles  in  (1)  and  (2).  This  is,  accordingly,  the 
general  expression  for  all  the  angles  which  have  the  same  sine 
as  A.  If  radian  measure  is  used,  and  XOP  =  a,  then  (3)  takes 
the  form 

Wir  +  (-ira.  (4) 

The  result  may  be  thus  expressed : 

sin  J.=sin  \n  -  1S0° -{- (-ly  A] ,  sin  «=sin  lmr+(-iya\.     (5) 

Since  cosec  6  = ,  the  general  expression  for  all  angles  which 

sin^ 
have  the  same  cosecant  is  the  same  as  the  general  expression  for  all 

angles  which  have  the  same  sine. 

EXAMPLES. 

1.  Find  an  expression  to  include  all  angles  which  have  the  same  sine  as 
135°. 

By  (3),  (4),  the  expression  is  n  •  180°  +  (-  1)^  135°,  or  mr  +  (-  1)"^- 


148 


PLANE  TRIGONOMETRY, 


[Ch.  XL 


2.   Find  the  general  value  of  the  angle  whose  sine  is  H Give  the  four 

1  V2 

least  positive  angles  which  have  sines  equal  to  H 

V2 

The  least  angle  whose  sine  is  -\ is  45°,  i.e.  7-      Hence  the  general 

V2  4 


value  is 


n  .  180° +  (-!)»» 45°,  i.e.  W7r+(-l)' 


To  find  the /owr  least  positive  angles,  put  n  =  0,  1,  2,  3,  in  this  expression. 

This  gives  45°    135°,  405°,  495°,  i.e.  -,  -tt,  -tt,  —  tt.      These  four  angles 

4    4       4        4 
can  also  be  obtained  by  means  of  a  figure. 

3.   Given  that  sin  6  =  ^ ;  find  the  general  value  of  d,  and  find  the  four 
least  positive  values  of  6. 


4.    As  in  Ex.  3  when  sin  d=—l. 


5.  As  in  Ex.  3  when  sin  ^=.95372. 


6.    As  in  Ex.  3  when  sin  e=. 39741.     7.  As  in  Ex.  3  when  sin  6=  -.57838. 


86.  General  expression  for  all  angles  which  have  the  same  cosine. 

Let  c  be  the  given  value  of  the  cosine.  It  is  required  to  find 
an  expression  to  include  every  angle  whose  cosine  is  c.  All  the 
angles  that  have  c  for  a  cosine  can  be  represented  geometrically, 
as  shown  in  Figs.  86,  87.  In  Fig.  86,  c  is  positive;  in  Fig.  87, 
c  is  negative. 

P  P^ 

1 


Xj_    o 


1 

Fia.  86. 


^1 


c^^-^0    X 

1 

Fia.  87. 


Let  XOP  be  the  least  positive  angle  whose  cosine  is  c,  and  let 
XOP  =  A  (in  degree  measure)  =  a  (in  radian  measure).  Angle 
XOPi  —  —  A  =  —a,  also  has  its  cosine  equal  to  c.  All  angles 
whose  terminal  line  is  OP,  have  cosines  equal  to  c.  All  these 
angles  are  included  in 

n  .  360°  +  A,  i.e.  2  titt  +  a,  (1) 

in  which  n  denotes  any  positive  or  negative  whole  number.  Also, 
all  angles  whose  terminal  line  is  OPi,  have  cosines  equal  to  c. 
All  these  angles  are  included  in 

n  '  360°  -  A,  i.e.  2  nr  -  a,  (2) 


86-87.]  GENERAL   VALUES.  149 

n  being  as  before.  Both  the  expressions,  (1),  (2),  are  evidently 
included  in 

n  •  360°  ±A,  or  2  nir  ±  a,  (3) 

in  which  n  is  any  positive  or  negative  whole  number.  Hence  (3) 
is  the  general  expression  for  all  angles  which  have  the  same 
cosine  as  A  or  a.     The  result  may  be  thus  expressed : 

cos  A  =  cos  (w  •  360°  ±  A)-,  cos  a  =  cos  (2  mr  ±  a).  (4) 

Since  sec  6  = ,  the  general  expression  for  all  angles  wJiich 

have  the  same  secant  islh^same  as  the  general  expression  for  all 
angles  which  have  the  same  cosine. 


EXAMPLES. 

1.  What  is  the  general  vakie  of  the  angles  which  have  the  cosine,  —  |  ? 
Give  the  three  least  positive  angles. 

The  least  positive  angle  whose  cosine  is,  —  |,  is  120°.  Hence,  the  general 
value  is,  by  (3),  n  -  360°  ±  120°,  i.e.  2  wtt  ±  fTr.  On  putting  n  =  0  and  1, 
the  three  least  positive  angles  are  found  to  be  120°,  360^  -  120°,  or  240°, 
360  4-  120°,  or  480°.  These  three  angles  may  also  be  found  by  means  of  a 
figure. 

2.  Given  that  cos  6  = :  find  the  general  value  of  6,  and  find  the  four 

least  positive  values  of  6. 

3.  As  in  Ex.  2  when  cos  6  =  .99106.       4.  As  in  Ex.  2  when  cos  d=  .46690. 
5.  As  in  Ex.  2  when  cos  ^=  -.72637.     6.  As  in  Ex.  2  when  cos  6=  -.40141. 

87.   General  expression  for  all  angles  which  have  the  same  tangent. 

Let  t  be  the  given  value  of  the  tangent.     It  is  required  to  find  an 
expression  to  include  all  angles  which  have  the  same  tangent  t. 

'P  p 

-\~^^^^   r"S^  x^  M  -1  o^^^i  Ux 

p  p 

^        Fig.  88.  -Fig.  89.  ^ 


^1 


All  the  angles  which  have  the  same  tangent  t  can  be  represented 
geometrically  as  in  Figs.  88,  89.  In  Fig.  88,  the  tangent  t  is 
positive,  in  Fig.  89,  it  is  negative. 


150  PLANE  TRIGONOMETRY.  [Ch.  XI. 

Let  XOP  =  A  (in  degrees)  =  a  (in  radians).     Then 

XOPi  =  180°  +  A  =  7r  +  a. 

Each,  angle  which  has  either  OF  or  OPi  for  its  terminal  line, 
has  its  tangent  equal  to  t.  All  the  angles  which  have  OP  for  a 
terminal  line  are  included  in  the  expression  m  •  360°  +  A,  that 
is,  in 

2m.l80°  +  ^  or  2m7r  +  a,  (1) 

in  which  m  denotes  any  positive  or  negative  whole  number. 

All  the  angles  which  have  OPi  for  a  terminal  line  are  included 
in  the  expression  m  •  360°+  (180°  +  A),  that  is,  in 

(2m  +  l)180°  +  ^,  or  (2 m -\- 1) tt  +  a.  (2) 

Both  these  sets  of  angles,  (1)  and  (2),  are  included  in  the 
expression 

n  •  180°  +  A,  or  nir  +  a,  (3) 

in  which  n  denotes  any  positive  or  negative  whole  number. 
Hence  (3)  is  the  general  expression  for  all  angles  which  have  the 
same  tangent  as  A  or  a.     The  result  may  be  thus  expressed: 

tan  A  =  tan  (n  •  180°  +  A) ;  tan  a  =  tan  (n-n-  +  a).  (4) 

Since  cot  0  = -,  the  general  expression  for  all  angles  ivliich 

have  the  same  cotangent  is  the  same  as  the  general  expression  for  all 
angles  which  have  the  same  tangent. 

EXAMPLES. 

1.  Find  the  general  value  of  6  when  tan  ^  =  1.     The  least  positive  angle 

whose  tangent  is  1,  is  ^.    Hence  6  =  rnr  -{--,  in  which  n  is  any  positive  or 

4  4 

negative  whole  number. 

Find  the  general  value  of  6,  and  the  four  least  positive  values  of  6  when  : 

2.  tan  6=  Vs.  3.    tan  6  =  .36727.  4.    tan  e  =  2.2998. 

5.   tan  0=. 71769.  6.    tan  0  =  - .90040.  7.    tan^  =  - 2.6511. 

8.  Find  the  general  expression  for  all  angles  which  have  the  same  sine 
and  cosine. 


sa.]  tnvEEst:  rntGoNoMETmc  functions.  151 

88.  Inverse  trigonometric  functions.  It  has  been  seen  that,  on 
the  one  hand,  the  value  of  the  sine  depends  on  the  value  of  the 
angle,  and,  on  the  other  hand,  the  value  of  the  angle  depends 
on  the  value  of  the  sine.  If  the  angle  is  given,  the  sine  can  be 
determined ;  if  the  sine  is  given,  the  angle  can  be  expressed. 
Hence,  on  the  one  hand,  tJie  sine  is  a  function  of  the  angle,  and, 
on  the  other  hand,  the  angle  is  a  function  of  the  sine.  The  latter 
function  is  said  to  be  the  inverse  function  of  the  former.  The 
same  holds  in  the  case  of  each  of  the  other  trigonometric  func- 
tions. Inverse  functions  are  usually  denoted  by  the  symbol 
described  below. 

The  two  statements  :  the  sine  of  the  angle  6  is  m,  (1) 

$  is  the  angle  whose  sine  is  m,  (2) 

are  briefly  expressed :  sin  0  =  m,  (3) 

6  =  sin~^  m.  (4) 

The  symbols  sin~^  m,  cos~^  m,  tan~^  m,  •••,  are  called  inverse  trigo- 
nometric furictions,  or  anti-trigonometric  functions,  or  inverse  circu- 
lar functions.  The  symbol  "  sin~^  m  "  is  read,  "  angle  whose  sine 
is  7/1,"  "anti-sine  of  m,"  "inverse  sine  of  m/'  "sine  minus  one  m." 
It  should  be  carefully  remembered  that  here,  —  1  is  not  an  alge- 
braical exponent,  but  is  merely  part  of  a  mathematical  symbol ; 

sin~^m  does  not  denote  (sin  m)~^,  that  is, ;  sin~^m  denotes  each 

sinm 
and  every  angle  whose  sine  is  m.     The  trigonometric  functions 
are  pure  numbers;  the  inverse  circular  functions  are  angles,  and 
are  denoted  by  the  number  of  degrees  or  radians  in  these  angles. 

For  instance,  if  ^  =  -  in  (3),  then  m  =  4-  ---: 
4       ^^'  ^V2 

if  m  =  +  —  in  (4),  then  6  =  sin"/— i— V  mr  +  (- 1)"- 

'     V2  V+V2;  4 

=  n.l80°  +  (-l)«45°, 
in  which  n  is  any  whole  number.     This  example  illustrates  what 
has  already  been  noted  in  Arts.  42,  43,  78,  namely : 

For  a  given  value  of  the  angle  0,  sin  ^  or  m  has  a  single  definite 
value. 

For  a  given  value  of  the  sine  m,  sin~^m  or  0  has  an  infinite 
number  of  values. 


152  PLAICE  TBIGONOMETUY.  .   [Ch.  XI. 

The  same  is  the  case  with  each  of  the  other  inverse  trigono- 
metric functions.  Thus  the  trigoiioiiietric  functions  are  single- 
valued,  and  the  inverse  circular  functions  are  multiple-valued. 

For  example,  if  cos  ^  =  — ,  then  B  -  cos'i  —  =  2  wtt  ±  ^(Ex.  2,  Art.  86); 

if  ^  =  tan-i  1,  then  ^  =  wtt  +  -  (Ex.  1,  Art.  87),  in  which  n  denotes  any- 
whole  number.  The  smallest  numerical  value  of  an  inverse  trigonometric 
function  is  called  the  principal  value  of  the  inverse  function.  For  instance, 
the  principal  value  of  sin"^^  is  30°,  of  tan-i  (—  1)  is  —  45°,  of  cos"^  (—  i) 

is  120°,  of  sin-i  (  -— )  is  -  60°. 

Note  1.  In  some  books  the  symbols  arc  sin  x,  arc  cos  x,  arc  tan  ic,  •••, 
are  used  for  inverse  trigonometric  functions.  These  symbols  are  read,  "arc 
sineic,"  ••-.     The  derivation  of  these  names  is  apparent  from  Art.  79. 

Note  2.  Algebraic.  If  ?/  is  a  function  of  a;,  say  /(x),  then  x  also  de- 
pends on  ?/,  and  hence,  is  some  function  of  y.  This  function  of  y  is  called 
the  inverse  function  of  f{x)  or  y,  and  is  usually  denoted  by  f~^(y).  For  in- 
stance, if  y  =f{x)  =  aj2,  then  x  =f-'^(y)  =  ±  Vy, 

It  will  be  observed  in  this  simple  example  that,  while  the  function  of 
X  has  a  single  value,  the  inverse  function  has  two  values.  In  other  words, 
?/  is  a  single-valued  function  of  x,  and  x  is  a  two-valued  function  of  y.  As 
shown  above,  it  y  =  sin  x,  then  x  =  sin~i  y  ;  y  is  ^  single-valued  function  of 
X,  but  X  is  a  multiple-valued  function  of  y. 

It  appears  from  Notes  1,  2,  that  the  English  notation  for  inverse  trigono- 
metric functions  avoids  the  old  geometrical  conceptions  of  trigonometric 
functions,  and  is  also  more  general  in  character.  The  inverse  trigonometric 
functions  are  frequently  met  in  calculus  and  applied  mathematics. 

89.  Sum  and  difference  of  two  anti-tangents.  Exercises  on  in- 
verse functions. 

Find  tan"^  m  -\-  tan"^  n,  and  tan"^  m  —  tan~^  n. 

Let  X  =  tan-^  m,   and  y  =  tan"^  n. 

Then  tan  x  =  m,   tan  y  =  n. 

Now  tm(x  +  y)=   tan x  + tan?/      (^rt.  51)     =  "^  +  "  • 
1  —  tan  X  tan  y  1  —  m7i 

,',x-}-y  =  tan-^  ^^  +  ^  ;  i.e.  tan"!  m  +  tan-^  n  =  tan-i  J>L±JL.  (1)  1 
1  —  7nn  1  -  nm  ' 

In  a  similar  manner  it  can  be  shown  that 

tan-i  m  -  ten"!  n  =  tan  >  ~^^  =i^.  (2) 

1   '  inn 


30.]  EXAMPLES,  153 

EXAMPLES. 
Kl.   Find  tan-i2  ±  tan-i|.     (Compare  Ex.  1,  Art.  51.) 

tan-i  2  +  tan-i  i  =  tan-i    ^  +  ^    =  tan-i  7  =  w  .  180°  +  81°  52'  11".5. 
1  —  2  .^ 

tan-i 2  -  tan-i 4  =  taii-i    ^  ~^    =  tan"!  1  =  ?i7r  +  -. 
'  1  +  2. i  ^4 

By  the  tables,  taking  acute  angles  only,  tan-i  2  =  63°  26'  4". 3,  tan-i  i 
=  18°  26'  6",  the  sura  is  81°  52'  10". 3,  and  the  difference  is  44°  59'  58". 3.  The 
slight  discrepancy  between  the  results  obtained  by  the  two  methods  is  due  to 
the  fact  that  the  angles  found  by  the  tables  are  only  approximately  correct. 

In  the  following  examples  test  or  verify  the  result  in  the  manner  shown 
in  Ex.  1. 

2.   Find  tan-i  7  ±  tan-i  3.  3.   Find  tan-i  2  +  tan-i .  5. 

4.   Find  tan-i  |  +  tan-i  \.  5.   Find  tan-i  3  +  tan-i  2  +  tan-i.6. 

(Suggestion.   Find  tan-i  3+tan-i  2,  then  combine  the  result  with  tan-i  .6.) 

6.  Find2tan-il.5,  2tan-i3,  2tan-i2,  3tan-i.2. 

7.  Show  that  2  tan-i  m = tan-i  -^-B- .    Show  that  2  ^ = tan-i  {  ^^^"^  \ . 

l-m2  Vl-tan2^/ 

8.  Show  that  4  tan~i  \  —  tan-i  -^\^  =  ^   when  the  angles    are   between 
0°  and  90°.  ^ 

9.  Find  sin  (sin~i  \  +  sin-i  \)  when  the  angles  are  between  0°  and  90°. 
10.    When  the  angles  are  between  0°  and  90°,  show  that : 


(a)  sin  (sin-i  m  ±  sin"i  n)  =  wVl  —  n^  ±  ny/1  —  w^. 

(Suggestion.     Let  x  =  sin-i  ?w,  y  =  sin""i «.) 


(6)  cos  (sin-i  m  ±  sin-i  n)  =  Vl  —  n^  Vl  —  m^  =F  ww. 
(c)   sin  (sin-i  m  ±  cos-i  n)  =  mw  i  Vl  —  m*  Vl  —  n^. 


(d)  cos  (sin-i  m  ±  cos"i  w)  =  ?z  Vl  —  m^  +  m  Vl  —  w^. 

11.  Find  sin  (sin-i  |  +  sin-i  f ),    cos  (sin~i  |  —  cos-i  |), 

sin  (cos-i  f  —  cos-i  j%),   sin  (tan~i  4  —  cos-i  |),    tan  (sec-i  3  —  sin-i  |), 
(a)  when  the  angles  are  between  0°  and  90°,  (6)  when  this  restriction  is  not 
imposed. 

12.  Two  lines,  AB,  AC,  intersect  a  horizontal  line  at  B,  C,  making  angles 
whose  tangents  are  f,  f.     Find  the  angle  BAC. 

13.  Two  lines,  LM,  LN,  make  angles  whose  tangents  are  |,  2,  with  a 
horizontal  line.     Find  the  angle  MLN. 


154  PLANE  TRIGONOMETBT,  [Ch.  XI. 

90.  Trigonometric  equations.  Trigonometric  equations  have 
appeared  in  many  of  the  preceding  articles.  When  an  angle,  6 
say,  is  the  unknown  quantity  in  a  trigonometric  equation,  the 
complete  solution  is  the  general  value  of  0  which  satisfies  the  equor 
tion.  For-  example,  if  a  be  an  angle  whose  sine  is  s,  then  the 
solution  of  the  equation, 

sin  B=zs,  that  is,  of  ^  =  sin-^s, 

is  6  =  nir  -\-  {— ly  a,  n  being  any  integer. 

EXAMPLES. 

(See  the  definition  of  principal  value  in  Art.  88.) 

V| 
2  * 


1.    Solve  the  equation  cos  d  = 


The  principal  value  of  6  is  -.    Hence  the  complete  solution  is  0  =  2  wtt  -jl  ^, 
6  6 

n  being  any  integer. 

2.  Solve  the  equation 


(See  Ex.  2,  Art.  86.) 

^  =  tan -11. 

0  =  n7r+I. 

(See  Ex.  1,  Art.  87.) 

The  principal  value  is  -. 

3.  Solve  the  equation        sin x cos x=—  \  V3.    . 

.♦.    sin X  Vl-sin^x  =  -  ^Vs.  .-.  sin2 x{l~  sin2 ic )  =  j%. 

.-.    sin*  X  -  sin2  a;  +  ^^  =  0.  .-.  (sin2  ic  -  |)  (sin2  aj  -  ;^)  =  0. 

.-.  sin2x  =  f ;  sm'^x  =  \. 

Whence  (a)  sin  a;  =  ±  V| ;  (6)  sin  a;  =  ±  ^. 

The  given  equation  shows  that  sinx  and  cosx  have  opposite  algebraic 
signs.     Hence,  x  can  only  be  in  the  second  and  fourth  quadrants. 

.-.  In  (a),  a;  =  120°,  300°,  etc.,  its  general  value  is  n  .  180  -  60°,  where  n 
is  any  positive  integer. 

In  (&),  x  =  150°,  330°,  etc.  ;   its  general  value  is  « •  180°  -  30°,  n  being 
any  positive  integer. 

4.  Solve  the  equation     sin  5  ^  +  sin  ^  =  sin  3  d. 

^2  sin  3  e  cos  2  6]=:  sin  3  0.     .-.  sin  3  ^(2  cos  2  ^  -  1)  =  0. 
~  .-.  (a) "sin  3^=0,     (6)  2  cos  2  0  -  1  =  0. 
From  (a),  3  ^  =  0°,  180°,  etc.  ;    the  general  value  of  3  ^  is  rnr  {n  being 
any  integer). 

.'.  e  =  0°,  60°,  etc.  ;  the  general  value  of  3  ^  is  ^. 

3 

From  (&),  Cos  2  ^  =  J.     .-.  26  =  ±  60°,  etc. ;  its  general  value  is  2  nv  ±  ^. 

o 

/.     6  =  ±  30°,  etc. ;  its  general  value  is    nr  ±  -• 

6 


90.] 


EXAMPLES. 


Find  solutions  of  these  equations  ; 

5.  3(sec2^  +  cot2^)=:  13. 

7.  seco;  +  tanx  =  2. 

9.  sec2  X  —  tan  x  =  3. 

11.  2sin  X  +  5  cosx  —  2. 

13.  4sin^cos2^=  1. 

15.  3  (tan2  q  ^  cot2  d)  =  10. 


6.  cot  ^  —  tan  ^  =  2. 

8.  sec2x  +  tanx  =  7. 

10.  cos  0  —  cos  7  ^  =  sin  4  6. 

12.  sin  2  ^  +  sin  4  ^  =  V2  •  cos  6. 

14.  tan*^  -  4  tan2  J.  +  3  =  0. 

16.  cos-i  X  —  sin-i  x  =  cos-i  x  \/3. 


N.B.     Questions  and  exercises  suitable  for  practice  and  review  on  the 
subject-matter  of  Chapter  XI.  imll  be  found  at  pages  197-199. 


CHAPTER  XII. 

MISCELLANEOUS   THEOREMS  AND   EXERCISES. 

91.  Chapters  II.-VIII.  were  devoted  to  the  oldest  and  the 
simplest  application  of  trigonometry;  namely,  the  measurement 
of  triangles.  Angles  and  the  trigonometric  functions  connected 
with  angles  were  more  fully  discussed  in  Chapters  IX.-XI.  This 
chapter  does  not  introduce  any  new  principles.  Most  of  its 
articles  may  be  regarded  as  exercises  on  the  relations  shown  in 
Chapters  II.-VIII.,  and  more  especially  on  the  properties  an- 
nounced in  Arts.  44,  50-52.  The  articles  just  mentioned  should 
be  reviewed.  Some  of  the  results  in  the  exercises  in  this  chapter 
are  useful  and  important;  but  the  student  should  direct  attention 
mainly  to  the  methods  whereby  the  results  are .  obtained,  so  that  he 
can  proceed  quickly  and  confidently  to  the  solutions  of  similar 
exercises.  These  solutions  require  a  ready  and  an  accurate 
knowledge  of  (that  is,  an  intelligent  familiarity  with)  the  for- 
mulas deduced  in  the  earlier  chapters.  It  is  on  this  account,  per- 
haps, that  such  exercises  are  regarded  with  favor  by  examiners. 

92.  Functions  of  twice  an  angle.     Functions  of  half  an  angle. 

Relations  (6)-(8),  Art.  60,  (3),  Art.  51,  give  the  sine,  cosine,  and  tangent 
of  twice  an  angle  in  terms  of  the  functions  of  the  angle.  On  rearranging  (7), 
(8),  Art.  50,  there  is  obtained, 


^U,A  =  ^ll=JfAA^    eoiA  =  yl^  +  'l^^^ 


(1) 


whence,  tan  ^  =  ™4  =a/M^4  (2) 

COS  A      '1  +  cos  2^. 

On  putting  I  x  for  A,  these  relations  take  the  forms 

(«)  sinix=^l^l|^,  (6)  cosix=V^^^' 

(c)  tan  ^  a;  =  Ji_ZL22i^ .  (3) 

><  1  +  cos  X 
In  (1),  (2),  (3),  angles  A  and  x  denote  any  angles. 

156 


91-93.]  FUNCTIONS  OF  THEEE  ANGLES.  157 

EXERCISES. 

1.  Account  for  the  ambiguity  of  the  radicals  in  (1),  (2),  (3). 

2.  Find  sin  45°,  given  tliat  cos  90°  =  0. 


From  (3)  a,  sin  45°  ^Jl-^os90"  ^  _1_. 


V2 

3.   Find  sin  22°  30',  cos  22°  30',  tan  22°  30'  by  means  of  (1),  (2).     Com. 
pare  the  values  vv^ith  those  given  in  the  tables. 

93.   Functions  of  three  times  an  angle.     Functions  of  an  angle  in 
terms  of  functions  of  one-third  the  angle. 

To  express  tan  S  A  in  terms  of  tan  A.     Let  A  denote  any  angle. 
tan3^  =  tan(2^  +  ^) 

^   .  .       z ^— r  +  tan  A 

tan  2A-\-  tan  A       1  —  tan^  A  {X  t  ni  \ 


l-tan2^tan^         ^_    2tan2^ 


l-tan^^ 


...  tan  3  ^  =  BtanA-Un^A^  qx 

1-3  tan2  A  ^ 

On  putting  x  for  ZA,  (1)  becomes 

^^^^^^3tanJ^^-tan3j^, 
1  -3tan2ix 

To  express  sin  3  A  in  terms  of  sin  A, 
sin  3  ^  =  sin  (2  ^  +  J.)  =  sin  2  ^  cos  J.  +  cos  2  ^  sin  A       [Art.  50,  (1)] 
=  2  sin  J.  cos2  ^  +  (1  —  2  sin^  A)  sin  A 
=  2  sin  ^(1  -  sin2  ^)  +  (1  -  2  sin2  A)  sin  A. 

.-.  sin  3  ^  =  3  sin  ^  -  4  sin^  A.  (2) 

In  a  similar  way,  cos  3  A  can  be  expressed  in  terms  of  cos  A. 

COS  3  -4  =  4  cos^  ^  -  3  COS  A.  (3) 

EXERCISES. 

1.  Derive  formula  (3). 

2.  On  substituting  x  for  3^1,  write  (2),  (3). 

3.  Express  formulas  (1),  (2),  (3),  and  the  results  of  Ex.  2,  in  words. 

4.  Assuming  the  value  of  sin  30°,  calculate  sin  90°. 


158  PLANE  TRIGONOMMBT.  [Ch.  Xll. 

5.  From  cos  30°,  derive  cos  90°  ;  from  tan  30°,  derive  tan  90°. 

6.  Derive    sin  180°,    cos  180°,    tan  180°,    from   sin  60°,    cos  60°,    tan  60°, 
respectively. 

7.  Derive  sin  75°,  cos  75°,  tan  75°,  from  sin  25°,  cos  25°,  tan  25°,  respec- 
tively, as  given  in  the  tables. 

8.  Derive  sin  37°  30',  cos  37°  30',  tan  37°  30',  from  the  ratios  of  75°. 
94.  Functions  of  the  sum  of  three  angles. 


i5Ilii±*2£A+tanO 
1  —  tan  A  tan  B 

^        tan  A  +  tan  B      .      ^ 

1 ^ •  tan  G 

1  —  tan  A  tan  B 


(1) 


_  tan^+tan  JB+tanO— tan^tanjBtan  (7 
1 —tan  A  tan  ^— tan  B  tan  (7— tan  C  tan  A 

Cor.  1.   It  Az=B  =  C,  (1)  reduces  to  (1),  Art.  93. 

CoR.  2.  If  A  +  B  +  C  =  180°,  then  tan(^  +  B  +  C)  =  0,  and,  accord- 
ingly, the  numerator  of  (1)  is  equal  to  zero.  Hence,  if  A,  B,  G,  are  the  three 
angles  of  a  triangle, 

tan  ^  +  tan  ^  +  tan  C  =  tan  A  tan  B  tan  C.  (2) 

CoR.  3.  If  A-\-  B  +  C  =  90°,  then  tan(^  +  JB  +  C)  =  oo,  and,  accord- 
ingly, the  denominator  of  (1)  is  equal  to  zero.     Hence, 

tan  ^  tan  ^  +  tan  5  tan  C  +  tan  a  tan  ^  =  1,  when  A  +  B  +  0  =  90°.     (3) 

EXERCISES. 

1.  Show  that  sin(^  +  B  -]-C)  =  smA  cos  J5  cos  C  +  cos  A  sin  S  cos  C  + 
cos  A  cos  J5  sin  C  —  sin  A  sin  B  sin  C. 

If  A  +  B  +  C  =  180°,  the  first  member  is  zero.  Division  of  the  second 
member  by  cos  A  cos  B  cos  C  will  give  relation  (2)  above. 

2.  Show  that  cos(^  +  J5  +  O)  =  cos  ^  cos  J5  cos  C  —  cos  A  sin  J5  sin  O  — 
sin  A  cos  i?  sin  C  —  sin  A  sin  B  cos  C.     What  does  this  become  when 

A-{-B  +  C=  180°  ? 

If  A  +  B  +  0=  90°,  the  first  member  is  zero.  Division  of  the  second 
member  by  cos  A  cos  B  cos  C  will  give  relation  (3)  above. 


94-95.]  IDENTITIES.  159 

3.    It  A  +  B-\-C=:  180°,  prove  that 

A         Ti        C 

COS  J.  +  COS  ^  +  COS  C  =  1  +  4  sin  —  sin  —  sin  -• 

2         2         2 

Since  A  +  B  +  G=1S0°,     A±A  =  QO°-^- 

COS  ^  +  COS  J5  +  cos  O  =  2  cos  ^^^  cos  ^-^  +  cos  C         [Art.  52  (7)] 

=  2  sin  -  cos  ^-^  +1-2  sin2  ^        [Art.  50  (7)] 

=  l  +  2sin^fcos^^:^-sin^'\ 
2V  2  2^ 

=  l  +  2sin^fcos^^l:^-cos^^il^l 
2V  2  2      y 

=  1 +  2sin  -  .2sin-sin-  [Art.  52  (8)] 

=  l  +  4sin^sin:5sin^. 
2         2         2 


4.  If  ^  +  ^  +  0  =  180°,  prove  that 

sin  -4  +  sin  5  +  sin  C 

5.  li  A-\-B-\-0-  180°,  prove  that 


ABC 

sin  ^  +  sin  5  +  sin  C  =  4  cos  —  cos  —  cos  — • 

2    2    2 


cos  ^  +  cos  JS  —  cos  0  =  —  1  +  4  cos  —  cos  —  sin  -. 

2         2        2 

6.  Also,  that  sin  (J.  +  B)  sin  (5  +  C)  =  sin  A  sin  C 

7.  Also,  that  sin2  A  +  sin2  B  +  sin2  O  =  2  +  2  cos  ^  cos  £  cos  C 

8.  If  ^  +  ^  +  O  =  90°,  show  that 

sin  2  ^  +  sin  2  5  —  sin  2  (7  =  4  sin  A  sin  B  cos  C. 

9.  Find  tan  4  ^4,  tan  5  A^  tan  6  J^,  tan  7  ^,  in  terms  of  tan  A. 

95.  Identities.  In  the  following  exercises  it  is  required  that 
the  first  member  be  changed  into  the  second  member.  When  it 
is  difficult  to  do  this,  help  is  sometimes  afforded  by  taking  some 
steps  in  changing  the  second  member  into  the  first.  The  direct 
steps  to  be  taken  from  the  first  member  to  the  second  may  be 
indicated  by  this  means.  No  general  directions  can  be  given  con- 
cerning the  making  of  these  transformations.  The  two  following 
suggestions,  however,  are  frequently  useful : 

(a)  Since  sin^^+  cos^^=l,  unity  can  be  substituted  for  the  first 
expression,  and  the  first  expression  can  be  substituted  for  unity. 
ip  (6)  The  change  of  tan  a;,  cot  x,  sec  ic,  cosec  x,  into  their  values 
ih  terms  of  the  sine  and  cosine,  is  sometimes  helpful. 

The  examples  in  Art.  hi  belong  to  this  class. 


160  PLANE  TRIGONOMETRY.  [Ch.  XII. 


EXERCISES. 

1.  Show  that  ^-^^^^^  =  tan2  A. 

1  +  cos  2  ^ 

1  -  cos  2  ^  ^  1  -(l-2sin2^)  ^  sin^^  _  ^^^^  ^ 
l  +  cos2^      l+(2cos2^-l)      cos2^ 

2.  Show  that  tan^  ^  ^  1  -  cos  2  ^^ 

1  +  cos  2  ^ 

tanM  =^^^'^  =  ^^^  -  cos 2^)  ^  1  -  cos 2^ 
cos2^     1(1  + cos  2^)      l  +  cos2^ 

Note.     The  fact  that  tan^  A  = ,  suggests  that  the  numerator  in  Ex.  1 

cos2^ 
be  expressed  in  terms  of  the  sine,  and  the  denominator  in  terms  of  the 
cosine.     In  Ex.  2,  the  plan  of  transformation  is  more  obvious. 

Prove  the  following  identities  : 

3.  _^^^!^_=sec2  5. 
2-sec2J5 

4.  l-2sin2(45°-^)=sin2A 

5.  cos2  A  -f  sin2  ^  cos  2  J5  =  cos2  B  +  sin2  B  cos  2  A. 

6.  1  +  cot  2  ^  cot  e  =  cosec  2  d  cot  d. 

7.  4  sin  A  sin  (60°  +  ^)  sin  (60°  -A)=  sin  3^. 

8.  cos  5  0  =  cos(3  ^  +  2  ^)  =  16  cos^  0-20  cos^  0  +  5  cos  6, 

9.  sin  5  0  =  16  sin^  0-20  sin^  0  +  5  sin  0. 

10.  tan(45^  +  A)-  tan  (45°  -  ^)  =  2  tan  2  A. 

11.  cos*  5  -  sin*  5  =  cos  2  5. 

j2^   sinl^^^cos3^^2sin2^-l. 

sin  u4  +  cos  ^ 

13.  sinx  +  sin2x     ^^^^^^ 
1  +  cos  ic  +  cos  2  x 

14.  4(cos«  X  +  sin^  ic)  =  1  +  3  cos^  2  x. 

15.  sin  4  ^  =  4  sin  A  cos^  A  —  A  cos  A  sin^  ^. 

16.  cos  4  ^  =  1  -  8  cos2  ^  +  8  cos*  A. 

96.   For  an  acute  angle  of  0  radians,  cos  6  >  1 ,  sin  8  >  9—  — • 

2  4 

By  Art.  60,  (7),  cos  (9  =  1  -  2  sin^  t  by  Art.  83,  sin  ^  <  ^- 
Hence,  cos  ^  >  1  —  2  ( |  ] ,  i.e.  cos  0  >  1  - 1-  \ 


By  Art.  50,  (5), 

2"~2      --""2-2      --2C-^'4) 


sin0  =  2sin^cos?  =  2tan^cos2^=2tan^ 


96,  97.]    COMPUTATION  OF  TRIGONOMETRIC  FUNCTIONS.  161 
But  by  Art.  83,  tan^>^,  and  sin^<^. 


I '-©•!"■ 


Hence  sin  ^  >  -—  1 1  — f  ^  )  1 ;  t'.e.  sin  9  >  0  -  ^. 


97.  One  method  of  computing  the  trigonometric  functions.  A 
method  of  computing  the  trigonometric  functions  of  angles  which 
are  in  an  arithmetic  progression  having  the  common  difference  Z)", 
will  now  be  shown. 

sin  in  +  1)  D"  +  sin  (n  -  1)  i)"  =  2  sin  nZ)"  cos  B'\  [Art.  52,  (5).] 

.-.      sin  (n  +  1) D"  =  2  sin  nD"  cos  D"  -  sin {n  - 1)  U\  (1) 

Also         cos  i)"  =  Vl-sin^Z)". 

Hence,  if  the  sines  of  the  angles  D",  2  D",  3D",  up  to  nD"  be 
known,  then  sin  {n  +  1)  D"  can  be  computed  by  formula  (1).  The 
other  functions  can  be  derived  from  the  sine. 

The  functions  for  angles  from  0°  to  45°  will  serve  for  the  angles 
from  45°  to  90°,  since  the  ratio  of  an  angle  is  the  co-ratio  of  its 
complement.  When  the  functions  have  been  computed  for  angles 
up  to  30°,  the  computations  for  angles  greater  than  30°  can  be 
made  more  easily.     For,  if  A  is  an  angle  less  than  30°, 

sin  (30°  +  ^)  4-  sin  (30°  -  ^)  =  2  sin  30°  cos  A  =  cos  A. ' 

sin  (30°  +  ^)  =  cos  ^- sin  (30°-^).  (2) 

Similarly,    cos(30°-f  ^)=cos(30° -^)~sinA  (3) 

In  formula  (1)  suppose  that  D"  =  10",  and  let  its  radian  meas- 
ure be  denoted  by  B, 

Then  sin  10"  <^,  >^-j-  [Arts.  83, 96.] 

Since  180°  =  tt, 

10"  = ^^ =  3.1415926535  ^  000048481368  .••  radians. 

180  X  60  X  60  64800 

.-.  sin  10"  <. 00004848...,  >  [.00004848  ••.- ^  (.00004848  ...)3]. 
Hence,  to  12  places  of  decimals,  sin  10"  =  .000048481368. 
From  this,  sin  20"  can  be  found  by  (1) ;  then  sin  30",  then  sin  40",  and  so  on. 
The  functions  of  several  angles  can  be  found  independently  of 
the  method  just  shown.     Formulas  involving  these  angles,  and 


162  PLANE  TRIGONOMETRY.  [Ch.  XII. 

Euler's  and  Legendre's  verification  formulas,  may  be  ased  to  test 
the  accuracy  of  the  tables.  The  latter  formulas  are  (see  Exs.  7- 
10,  Ch.  XII.), 

sin(36°+^)-sin(36°-^)-sm(72°+^)  +  sm(72°-^)  =  sin^  (4) 
cos(36°H-^)+cos(36°-^)-cos(72°+^)-cos(72°-^)=cosA(5) 

EXERCISES. 

1.  Test  the  tables  of  natural  sines  and  cosines  by  means  of  formulas  (4), 
(5),  taking  A  equal  to  4°,  10°,  15°,  and  other  values.  2.  Assuming  the  func- 
tions of  1°  as  known,  calculate  the  sines  of  2°,  3°,  4°,  6°,  6°,  by  formula  (1). 
3.  By  means  of  formulas  (2).  (3),  calculate  the  sines  and  cosines  of  33°, 
37°,  41°,  47°,  53°,  67°,  and  other  angles. 

98.  Trigonometry  defined.  Branches  of  trigonometry.  Before 
concluding  this  text-book  it  may  be  well  to  indicate  to  the  student 
the  relation  of  the  part  of  trigonometry  treated  in  the  preceding 
pages  to  the  subject  as  a  whole,  and  also  to  try  to  give  him  a 
little  idea  of  another  branch  of  trigonometry ;  namely,  analytical 
trigonometry. 

In  Chapters  II.-IX.,  plane  angles,  the  solution  of  plane  tri- 
angles, and  applications  connected  therewith  were  discussed. 
This  is  what  is  usually  known  as  plane  trigonometry.  The  study 
of  solid  angles,  the  solution  of  spherical  triangles,  and  the  as- 
sociated practical  applications,  constitute  spherical  trigonometry. 
These  branches  of  mathematics  are  founded  on  geometrical  con- 
siderations, and  may  be  looked  upon  as  applications  of  algebra  to 
geometry.  Pure  mathematics  is  sometimes  regarded  as  consisting 
of  two  great  branches ;  namely,  geometry  and  analysis.  Analysis 
includes  algebra,  infinitesimal  calculus,  and  other  subjects  which 
employ  the  symbols,  rules,  and  methods  of  algebra,  and  do  not 
rest  upon  conceptions  of  space.  (Geometrical  ideas  may  be  used 
in  analysis,  however,  for  the  sake  of  exposition  and  illustration, 
and,  on  the  other  hand,  algebra  may  be  employed  in  expounding 
the  principles  of  geometry.)  Since  the  eighteenth  century,  trigo- 
nometry has  also  been  treated  as  a  branch  of  analysis.* 

*The  meaning  of  the  word  "  analysis"  thus  used  in  mathematics,  should 
not  be  confounded  with  the  ordinary  meaning  of  the  word,  or  with  the 
meaning  attached  to  the  term  *'  analysis  "  in  logic. 


98.]  ANALYTICAL    TRIGONOMETRY.  163 

Analytical  (or  algebraical)  trigonometry  treats  of  the  general 
relations  of  angles  and  their  trigonometric  functions  without  any 
reference  to  measurement.  It  discusses,  among  other  things,  the 
development  of  exponential  and  logarithmic  series,  the  connections 
between  trigonometric  and  exponential  functions,  the  expansions 
of  an  angle  and  its  trigonometric  functions  into  infinite  series,  the 
calculation  of  ir,  the  summation  of  series,  and  the  factorization  of 
certain  algebraic  expressions.  The  properties  stated  in  formulas, 
(l)-(3)  Art.  44,  (l)-(8)  Art.  50,  (l)-(8)  Art.  52,  (l)-(3)  Art.  93,, 
are  analytical  properties,  and  can  he  derived  without  the  did  of  geom- 
etry. Analytical  trigonometry  includes  hyperbolic  trigonometry ; 
that  is,  the  treatment  of  what  are  called  the  hyperbolic  functions. 

While  the  trigonometric  functions  may  be  defined  and  discussed 
on  a  geometrical  basis,  as  done  in  this  book  (and  this  is  the  easiest 
way  for  beginners),  it  may  be  stated  that  they  can  also  be  defined 
and  their  properties  deduced  on  a  purely  algebraic  basis.  It  is 
beyond  the  scope  of  this  work  to  show  this,  but  the  student  may 
obtain  a  little  light  on  the  subject  by  reading  Notes  A  and  D. 
It  may  be  stated  further,  that,  under  certain  restrictions,  some  of 
the  most  important  theorems  and  properties  found  in  analytical 
trigonometry  can  be  derived  easily  in  an  elementary  course  in  the 
infinitesimal  calculus.  It  has  been  pointed  out  that  the  trigono- 
metric functions  can  be  defined  in  a  purely  geometrical  manner, 
and  in  a  purely  algebraic  manner ;  they  can  also  be  given  defini- 
tions depending  on  the  infinitesimal  calculus,  and  their  properties 
deduced  therefrom.  Finally,  it  may  be  said  that  trigonometry  is 
merely  a  brief  chapter  in  the  modern  Theory  of  Functions,  and 
may  be  defined  as  the  science  of  singly  periodic  /mictions  (see 
Art.  78).  For  a  treatment  of  trigonometry,  either  as  a  part  of 
algebra,  or,  as  "an  elementary  illustration  of  the  application 
of  the  Theory  of  Functions,"  see  Lock,  Higher  Trigonometry ; 
Loney  (Part  II.),  Analytical  Trigonometry ;  W.  E.  Johnson,  Trea- 
tise on  Trigonometry,  Chaps.  XII.-XXII. ;  Casey,  A  Treatise  on 
Plane  Trigonometry ;  Levett  and  Davison,  Elements  of  Plane 
Trigonometry  (Parts  II.,  III.,  Eeal  Algebraical  Quantity,  Com- 
plex Quantity);  Hay  ward,  Vector  Algebra  a7id  Trigonometry; 
Hobson,  A  Treatise  on  Plane  Trigonometry ;  Chrystal,  Algebra, 
Part  I.,  Chap.  XII. ;  Part  II.,  Preface,  and  Chaps.  XXIX.,  XXX. 


APPENDIX, 


NOTE  A. 
HISTORICAL  SKETCH. 

The  most  ancient  mathematical  writing  known  at  the  present  time  is  an 
Egyptian  papyrus  preserved  in  the  British  Museum.  It  is  the  work  of 
Ahmes,  an  Egyptian  priest  who  lived  at  least  seventeen  hundred  years  b.c, 
and  is  believed  to  have  been  founded  on  older  works  dating  as  far  back  as 
3400  B.C.  The  treatise  is  concerned  with  practical  mathematics,  and  merely, 
gives  rules  for  making  geometrical  constructions  and  determining  areas. 
The  area  of  an  isosceles  triangle  is  obtained  by  taking  the  product  of  half 
the  base  and  one  of  the  sides.  The  area  of  a  circle  is  found  by  deducting 
from  the  diameter  one-ninth  of  its  length,  and  squaring  the  remainder- -a 
proceeding  which  is  equivalent  to  taking  tt  =  3.1604  ••-. 

The  ancient  Greeks  brought  geometry  to  a  high  state  of  perfection,  but 
showed  little  aptitude  for  algebra  and  trigonometry.  They  were  not  inclined 
to  be  satisfied  with  approximate  results,  and  regarded  the  practical  applica- 
tion of  mathematics  as  degrading  to  the  science.  Trigonometry  was  invented 
to  supply  practical  needs,  and  its  development,  in  the  earlier  stages,  was  due 
to  men  of  the  Egyptian,  the  Hindoo,  and  the  Semitic  races. 

Astronomy  was  one  of  the  studies  most  cultivated  by  the  ancients,  but 
astronomy  could  not  advance,  or  even  become  a  science,  without  the  aid  of 
trigonometry,  Hipparchus  of  Nicsea  in  Bithynia,  the  greatest  astronomer 
of  antiquity,  who  flourished  about  160-120  e.g.,  is  regarded  as  the  founder 
of  trigonometry,  which  he  developed  solely  as  a  necessary  part  of  astronomy. 
Moreover,  trigonometry  continued  to  exist,  for  the  most  part,  merely  as  a 
handmaid  of  astronomy  for  over  eighteen  hundred  years.  On  this  account,  the 
theorems  of  spherical  trigonometry  were  developed  earlier  than  those  of  plane 
trigonometry.  Of  the  writings  of  Hipparchus,  all  but  one  have  been  lost ; 
but  it  is  known  that  he  constructed  a  table  of  chords,  which  serves  the  same 
purpose  as  a  table  of  natural  sines.  Hero  of  Alexandria,  who  flourished 
some  time  between  155  and  100  b.c,  and  is  supposed  to  have  been  a  native 
Egyptian,  found  the  area  of  a   triangle    in  terms  of  its  sides,  and  placed 

166 


166  PLANE  TRIGONOMETRY. 

engineering  and  land-surveying  on  a  scientific  basis.  Ptolemy,  a  native  of 
Egypt,  the  records  of  whose  observations  cover  the  period  127-151  a.d.,  wrote 
the  Syntaxis  Mathematica  (called  the  Almagest  by  the  Arabs),  a  work 
founded  on  the  investigations  of  Hipparchus.  This  was  regarded  as  a  kind 
of  astronomical  Bible  for  thirteen  hundred  years,  until  the  Ptolemaic  theory, 
namely,  that  the  sun,  planets,  and  stars  revolve  around  the  earth,  was  shown 
to  be  erroneous  by  Copernicus  and  Galileo.  The  Almagest  is  divided  into 
thirteen  books.  Book  I.  treats  of  plane  and  spherical  trigonometry,  contains 
a  very  accurate  table  of  chords,  probably  derived  from  Hijjparchus,  and 
shows  the  method  of  forming  the  table.  It  develops  spherical  before  plane 
trigonometry,  and  does  not  give  the  solution  of  plane  triangles.  "  Whereas 
the  Ptolemaic  system  (of  astronomy)  was  .  .  .  overthrown,  the  theorems  of 
Hipparchus  and  Ptolemy,  on  the  other  hand,  will  be,  as  Delambre  *  says, 
forever  the  basis  of  trigonometry."  t 

Whatever  advance  was  made  in  trigonometry  during  the  thousand  years 
after  Ptolemy,  was  due  to  the  Hindoos  and  Arabs.  The  Hindoos  had  tables 
of  the  half-chords,  or  sines,  and  found  that  the  arc  equal  in  length  to  the 
radius  contained  3438'.  Aryabhatta  (476-530  a.d.  ?)  wrote  a  work  containing 
sections  on  astronomy,  spherical  and  plane  trigonometry.  This  contained 
tables  of  natural  sines  of  the  angles  in  the  first  quadrant  at  intervals  of  3f°, 
the  sine  being  defined  as  the  semi-chord  of  twice  the  angle.  Pe  gave  3.1416 
as  the  value  of  tt.  Other  writers  were  Brahmagupta,  born  598,  and  Bhaskara, 
about  1150,  who  gave  some  trigonometric  formulas.  The  Hindoos  knew 
how  to  solve  plane  and  spherical  right  triangles. 

During  the  period  of  the  Dark  Ages  in  Europe,  the  sciences  of  the  Greeks 
and  Hindoos  were  preserved,  and,  to  some  slight  extent,  improved  by  the 
Arabs.  The  latter  studied  trigonometry  only  for  the  sake  of  astronomy. 
The  term  sine  is  due  to  the  celebrated  Arabian  astronomer  Al  Battani 
(Albategniiis),  a  native  of  Syria,  who  died  about  930  a.d.  Another  Arabian 
astronomer,  Abii  H  Wafa  (940-998),  a  native  of  Persia,  was  the  first  to  intro- 
duce the  tangent  of  the  arc  into  the  science  ;  he  calculated  a  table  of  tangents. 
Among  the  Western  Arabs,  to  whom  the  development  of  the  subject  is  in- 
debted, were  Ihn  Yunos  of  Cairo  (died  1008),  and  Gabir  ben  AJlah,  who 
was  born  at  Seville  and  who  died  at  Cordova  in  the  latter  part  of  the  eleventh 
century.  The  latter  wrote  an  astronomy  in  nine  books,  the  first  of  which  is 
devoted  to  trigonometry  ;  he  also  contributed  to  the  advancement  of  spherical 
trigO]iometry. 

The  next  stage  in  the  history  of  trigonometry  is  marked  by  the  introduc- 
tion of  the  Arabian  works  into  Europe,  and  the  development  of  the  arithmet- 
ical part  of  the  subject,  especially  the  calculation  of  tables.     This  was  largely  ' 

*Jean  Baptiste  Delambre  (1749-1822),  a  French  mathematician  who  de- 
rived important  formulas  in  spherical  trigonometry. 


t  Ency.  Brit.,  Art.  Ptolemy. 


J 


APPENDIX,  167 

the  work  of  German  astronomers,  and  chiefly  of  Begiomontanus  and  Mheticus. 
Georg  Purbach  (1423-1461),  professor  of  mathematics  and  astronomy  at  the 
University  of  Vienna,  wrote  a  table  of  natural  sines  computed  for  intervals 
of  ten  minutes,  which  was  published  in  1541.  Begiomontanus  (John  Mtiller) 
(1436-1476),  a  native  of  Franconia,  who  was  one  of  the  greatest  mathemati- 
cians that  Germany  has  ever  produced,  in  conjunction  with  Purbach  made  a 
translation  of  the  Almagest,  which  was  published  in  1496.  In  this  he  sub- 
stituted sines  for  chords,  and  gave  a  table  of  natural  sines.  He  reinvented 
the  tangent,  and  made  a  table  of  natural  tangents  for  all  degrees  of  the  quad- 
rant ;  this  was  published  in  1490.  In  1464  he  wrote  his  De  Triangulis,  which 
was  the  earliest  modern  systematic  exposition  of  plane  and  spherical  trig- 
onometry. This  was  printed  in  1533,  and  a  second  edition  appeared  in  1561. 
The  only  functions  introduced  were  sines  and  cosines.  Copernicus  (1473- 
1543),  born  in  Prussia,  wrote  a  short  text-book  on  the  subject  about  1500, 
which  was  published  in  1542.  Bheticus  (Georg  Joachim)  (1514-1576),  a 
native  of  the  Tyrol,  professor  of  mathematics  at  Heidelberg,  constructed 
tables  (published  in  1596)  which  are  the  basis  of  those  still  in  use.  He  intro- 
duced secants  and  cosecants,  and  found  the  values  of  sin  2  6,  sin  3  0  in  terms 
of  sin  6,  cos  6.  Hitherto  the  trigonometric  functions  had  been  considered  as 
lines  related  to  circular  arcs.  Rheticus  was  the  first  who  constructed  the 
right  triangle  and  used  the  ratio  definitions  which  depend  directly  on  the 
angle.  These  definitions  were  not  adopted,  however,  and,  although  intro- 
duced two  hundred  years  later  by  Euler  in  1748,  they  did  not  come  into 
common  use  until  after  the  middle  of  the  present  century.  Pitiscus  (1561- 
1613),  professor  of  mathematics  at  Heidelberg,  made  important  corrections 
in  and  additions  to  the  tables  of  Rheticus.  His  trigonometry,  published  in 
1599,  contained  formulas  for  cos  (A±  B),  sin  (A  —  B).  Adrian  Bomanus 
(1561-1625),  a  Belgian  mathematician,  professor  at  the  University  of  Lou  vain, 
first  found  the  formula  for  sin  (yl  +  B).  Franrms  Vieta  (1540-1603),  the 
greatest  French  mathematician  of  the  sixteenth  century,  extended  the  tables 
of  Rheticus.  He  made  one  of  the  earliest  attempts  to  find  the  value  of  tt  by 
means  of  infinite  series,  and  was  the  first  who  made  any  considerable  appli- 
cation of  algebra  to  trigonometry.  In  his  work,  Ad  Angulares  Sectiones, 
he  gave  formulas  for  sin  nd,  cos  nO,  in  terms  of  sin  6,  cos  d.  John  Napier 
(1550-1617)  discovered  the  important  formulas  in  spherical  trigonometry 
which  are  commonly  called  Napier's  Analogies.  His  invention  of  logarithms 
greatly  lessened  the  arithmetical  work  necessary  in  astronomy  and  trigonom- 
etry, and  thus  ushered  in  a  new  era  in  the  history  of  these  sciences.  Edmund 
Ounter  (1581-1026),  professor  of  astronomy  at  Gresham  College,  London, 
gave  the  first  tables  of  logarithms  of  sines  and  tangents.  He  first  used  the 
terms  cosine,  cotangent,  cosecant.  Albert  Girard  (1590-1634),  a  Flemish 
mathematician,  published  a  trigonometry  in  which  the  contractions  sin,  tan, 
sec  were  used.  William  Oiightred  (1575-16(50),  an  English  mathematician, 
wrote  a  trigonometry,  published  in  1657,  containing  abbreviations  for  sine, 


168  PLANE  TRIGONOMETRY. 

cosine,  but  they  did  not  come  into  general  use  until  Euler  reintroduced  them 
nearly  a  century  later.* 

Thus  far,  trigonometry  had  been  confined  to  the  bounds  set  by  the  ancients, 
namely,  to  expressing  the  relations  between  the  sides  and  angles  of  plane  and 
spherical  triangles,  to  the  solution  of  triangles,  and  to  the  calculation  of 
tables.  Trigonometry  had  been  founded  on  geometrical  conceptions,  and 
was  regarded  mainly  as  an  appendage  of  geometry  and  astronomy.  In  the 
seventeenth  and  eighteenth  centuries,  however,  a  new  branch  of  the  subject, 
namely,  analytical  trigonometry,  was  created,  chiefly  by  the  genius  of  De 
Moivre  and  Euler.  In  the  new  development  of  the  science,  the  symbols, 
rules,  and  methods  of  algebra  were  employed,  and  geometrical  conceptions 
were  disregarded.  [See  Art.  98  and  Note  D.]  The  older  trigonometry  still 
retains  its  position  as  a  necessary  department  of  applied  mathematics.  The 
modern  analytical  (or  algebraical)  side  of  the  subject,  however,  has  been  so 
highly  developed  since  the  middle  of  the  eighteenth  century,  and  its  results 
are  so  much  employed  in  other  branches  of  mathematical  and  physical  sci- 
ence, that  it,  may  be  regarded  as  the  larger  and  more  important  part  of 
trigonometry. 

The  new  development  began  with  the  discovery  and  investigation  of  expo- 
nential, logarithmic,  and  trigonometric  series.  The  chief  investigators  of 
infinite  series  were  :  John  Wallis  (1G16-1703),  professor  of  geometry  at 
Oxford  ;  James  Gregory  (1638-1675),  professor  of  mathematics  at  Edin- 
burgh ;  Nicolaus  Mercator,  died  1687,  a  native  of  Holstein,  who  settled  in 
England;  Isaac  Newton  (1642-1727);  Gottfried  William  Leibnitz  (1646- 
1716).  Several  of  these  series  greatly  simplified  the  calculation  of  tt  ;  some 
of  them  were  obtained  by  means  of  the  infinitesimal  calculus  invented  by 
Newton  and  Leibnitz.  Before  1669,  Newton  obtained  the  series  for  the 
arc  in  powers  of  the  sine,  and  the  series  for  the  sine  and  cosine  in  powers 
of  the  arc.  In  1670,  Gregory  discovered  the  series  for  the  arc  in  powers 
of  the  tangent,  and  the  series  for  the  tangent  and  secant  in  powers  of  the 
arc  ;  Leibnitz  discovered  the  first  of  these  independently  in  1673. 


*  "  To  England  falls  the  honour  of  having  produced  the  earliest  European 
writers  on  trigonometry."  (Cajori,  History  of  Mathematics,  p.  135.) 
Thomas  Bradwardine  (1290  P-1349) ,  archbishop  of  Canterbury,  Richard  of 
Wallingford  (1292?-1336),  abbot  of  St.  Albans,  John  Mauduith  (about 
1310),  fellow  of  Merton  College,  Oxford,  who  were  mathematicians  and 
astronomers,  left  writings  containing  trigonometry  and  tables  drawn  from 
Arabic  sources.  The  earliest  English  books  in  which  spherical  trigonometry 
is  used,  are  those  of  Thomas  Digges  (died  1595),  one  of  the  foremost  English 
mathematicians  of  the  sixteenth  century.  The  earliest  book  in  which  plane 
trigonometiy  is  introduced,  is  a  -work  published  by  Thomas  Blundeville 
in  1594. 


J 


APPENDIX,  169 

John  Bernoulli  (1667-1748),  a  native  of  Switzerland,  originated  the  idea 
of  trigonometric  functions,  and  treated  trigonometry  as  a  branch  of  analysis. 
He  was  tlie  first  to  obtain  real  results  by  using  the  symbol  V—  1.  Abraham 
de  3Ioivre  (1667-1754),  a  French  Huguenot  who  settled  in  London,  did  much 
to  advance  analytical  trigonometry,  by  his  use  of  (so-called)  imaginary 
quantities,  and  the  discovery  of  the  great  fundamental  theorems,  which  are 
called  by  his  name.  (See  Note  D.)  Johann  Heinrich  Lambert  (1728-1777), 
a  native  of  Alsace,  developed  de  Moivre's  theorems,  introduced  the  functions 
called  hyperbolic  sine  and  cosine,  and  showed  their  connection  with  the 
hyperbola.  He  also  found  that  ir  is  incommensurable.  Modern  trigonometry 
is  indebted  most  of  all  to  Leonhard  Euler  (1707-1783),  a  native  of  Switzerland. 
In  his  Introductio  in  Analysin  Injinitorum,  published  in  1748,  he  system- 
atized and  generalized  what  was  then  known  about  algebra  and  trigonometry. 
He  discussed  the  expressions  of  functions  in  series,  and  treated  trigonometry 
as  a  branch  of  analysis.  The  latter  was  effected  by  regarding  trigonometric 
functions,  not  as  straight  lines  belonging  to  arcs,  and  thus  depending  on  the 
radius  of  a  circle,  but  as  ratios,  and  thus  as  functions  of  the  angle  only.  He 
reintroduced  the  abbreviations  now  used.  [This  was  done  simultaneously 
in  England  by  Thomas  Simpson  (1710-1761),  professor  of  mathematics  at 
Woolwich,  in  his  trigonometry,  also  published  in  1748.]  Euler  first  showed 
the  connection  between  exponential  and  trigonometric  functions  (see  Note 
D),  and  discovered  many  of  their  analytical  properties.*  Since  the  time 
of  Euler,  analytical  trigonometry  has  benefited  by  the  immense  advances 
made  in  the  theory  of  functions  of  complex  quantities ;  that  is,  quantities 
of  the  form  x  +  V^H!  y.  It  is  now  coming  to  be  regarded,  more  properly 
and  more  logically,  as  an  elementary  chapter  in  the  modern  theory  of  func- 
tions.    See  Chrystal,  Algebra^  Part  II.,  p.  vii.t 

NOTE   B. 

1.  Projection  definition  of  the  trigonometric  ratios.  [Supplementary  to 
Art.  40.]  In  Fig.  20,  Art.  28,  MN  is  the  projection  of  AB  on  LB,  and  NM 
is  the  projection  of  BA  on  LB.  In  naming  the  projection,  the  points 
obtained  by  projection  are  taken  in  the  same  order  as  the  corresponding 
points  in  the  original  line.  It  is  apparent  that,  for  any  line,  the  projections 
upon  a  series  of  parallel  lines  are  equal.  For  instance,  AD  =  MN,  Fig.  20. 
This  may  also  be  seen  by  drawing  a  series  of  lines  parallel  to  LB  and  pro- 
jecting AB  upon  them. 

*  The  first  English  book  in  which  trigonometry  received  an  analytical 
treatment  was  that  of  Robert  Woodhouse  (1773-1827),  professor  at  Oxford, 
which  was  published  in  1809. 

t  The  principal  sources  from  which  this  historical  sketch  has  been  drawn, 
are  Hobson,  Article  Trigonometry  (JEncyclopcedia  Britannica,  Oth  edition), 
Ball,  A  Short  History  of  Mathematics^  Cajori,  A  History  of  Mathematics, 


170 


PLANE   TRIGONOMETRY. 


Suppose  that  in  Fig.  36,  Art.  40,  YOY^  be  drawn  at  right  angles  to  X^OX. 
Then 

OM  is  the  projection  of  the  turning  line  OP  upon  OX, 

MP  is  equal  to  the  projection  of  the  turning  line  OP  upon  OY. 

In  two  cases  in  Fig.  36,  the  projection  of    OP  on  OX  is  in  the  direction 
opposite  to  OX,  that  is,  it  is  negative ;  in  two  cases,  the  projection  of  OP 
on  0  F  is  opposite  to  the  direction  of  Y,  that  is,  it  is  negative. 
The  definitions.  Art.  40,  may  now  be  stated  as  follows : 


.i^^^proJ-OPonOr      ^^^  j^^Vm^OPonOY  ^ 
OP  proj.  OP  on  OX 


sec-4=- 


OP 


proj.  OP  on  OX 
OP 


^^^^^p^oj^OPonOX.      eot^=PMJ>P^OX    eosec^=- 

OP  proj.  OP  on  O  r  proj.  OP  on  0  F 

These  differ  from  (1),  Art.  40,  merely  in  the  fact  that  names  are  given  to 
OM  and  MP.  The  properties  shown  in  Chap.  V.  follow  from  these 
definitions. 

2.  Theorem  on  projection.  The  projection  of  one  side  of  a  polygon  upon 
any  straight  line  is  equal  to  the  algebraic  sum  of  the  projections  of  the 
other  sides. 


s     K 


L     P 


FiQ.  91. 


Let  PQRS  be  any  polygon.    Draw  parallel  lines  P/),  Qq,  Bi\  Ss,  from  its 
vertices  to  any  straight  line  LK.     Then  it  is  apparent  that 


i.e. 


ps  =  pq  +  qr  +  rs  ; 
proj.  PS  =  proj.  PQ  +  proj.  QR  +  proj.  RS. 


In  Fig.  91,  ?*s,  the  projection  of  RS,  is  negative.  This  proposition  is  true 
whether  the  projection  be  oblique  or  orthogonal.  The  theorem  may  also  be 
stated  thus : 

The  projection  of  a  broken  line  upon  a  straight  line  is  equal  to  the  projec- 
tion of  the  line,  drawn  from  the  initial  point  to  the  terminal  point  of  the 
broken  line.  Thus,  the  projection  of  the  broken  line  PQRS  upon  any 
straight  line  is  equal  to  the  projection  of  PS  upon  the  same  line. 


APPENDIX.  171 

3.  The  sine  and  cosine  of  the  sum  of  two  angles.  [Supplementary  to 
Art.  46.] 

Let  the  construction  be  made  as  indicated  in  Art.  46.     Then 

cin^^a.  PN  _  Pr03»  Oi^ on  or _ proj.  0^  on  or     proj.QPonOr 

[Theorem  in  (2).] 
_  proj.  OQ  on  or    OQ     proj.  ^P  on  Or   QP 
OQ  'op  QP  '  OP 

=  sui  J.  cos  P  +  sin  VQP  sin  B 
=  sin  A  cos  B  +  cos  A  sin  B. 

coc  CA  I  B^  =  P^^J-  ^^  ^"  ^^=  P^^-^-  OQ^^O^  \  P^^oJ-  QP  o^^  ^-^ 
^  ^  OP  OP  OP 

_pro3.0(>onOX    0^     proj.  QP  on  OX    QP 
OQ  'op  QP  'op 

=  cos  A  cos  B  +  cos  VQP' sin  B 

=  cos  A  cos  B  —  sin  J.  sin  B. 

in  the  projection  proof  of  the  addition  formulas  for  the  sine  and  cosine,  A 
and  B  can  have  any  magnitudes,  positive  or  negative.  The  formulas  for 
sin(yl  —  B),  cos(^  —  jB),  can  also  be  derived  by  substituting  —  B  for  +  B 
In  the  addition  formulas. 

NOTE  C. 

[Supplementary  to  Arts,  9,  72.] 
ON  THE  LENGTH  AND   AREA   OF   A   CIRCLE. 

1.  The  main  purpose  of  this  note  is  to  outline  a  method  of  approximating 
to  the  value  of  ir ;  that  is,  to  the  ratio  of  the  length  of  a  circle  to  its  diameter. 
This  method  depends  only  on  elementary  geometry.*  There  are  simpler  and 
more  expeditious  methods  of  finding  rr,  but  they  require  a  greater  knowledge 
of  mathematics  than  beginners  in  trigonometry  generally  possess. 

By  the  methods  of  elementary  geometry,  as  shown  in  the  texts  of  Euclid 
and  others,  regular  polygons  of  3,  4,  5,  6,  15  sides  can  be  inscribed  in,  and 
circumscribed  about  a  given  circle.  Moreover,  inscribed  and  circumscribing 
regular  polygons  of  2,  4,  8,  16,  •••,  times  each  of  those  numbers  of  sides  can 
also  be  constructed  by  successively  bisecting  the  arcs  subtended  by  the  sides, 
and  joining  the  consecutive  points  of  division.     This  process  can  evidently  be 

*  A  section  on  the  mensuration  of  the  circle  is  given  in  many  geometries. 
Reference  may  be  made  to  the  geometries  of  Beman  and  Smith  (Ginn  &  Co,)^* 
Gore  (Longmans,  Green,  &  Co.),  Phillips  and  Fisher  (Harpers),  and  others. 


172 


PLANE  TRIQONOMETRT, 


carried  on  until  tlie  inscribed  and  circumscribing  polygons  have  an  infinitely 
great  number  of  sides  ;  that  is,  regular  polygons  of  3.2'*,  4.2«,  5.2",  15.2"  sides, 
n  being  any  positive  integer,  can  be  inscribed  in,  or  circumscribed  about,  a 
given  circle. 

2.  Outline  of  a  proof  of  the  theorem  that  the  lengths  of  circles  are  pro- 
portional to  their  diameters. 

(a)  The  length  of  a  circle  is  gi-eater  than  the  perimeter  of  an  inscribed 
polygon,  and  is  less  than  the  perimeter  of  a  circumscribing  polygon  of  any 
finite  number  of  sides. 

(&)  As  the  number  of  sides  of  a  regular  polygon  inscribed  in,  or  circum- 
scribed about,  a  circle  is  increased,  the  length  of  the  perimeter  of  the  polygon 
approaches  nearer  and  nearer  to  the  length  of  the  circle.  In  other  words,  by 
increasing  the  number  of  sides,  the  difference  between  the  length  of  the 
perimeter  of  the  polygon  and  the  length  of  the  circle  may  be  made  as  small 
as  one  please,  and  this  difference  approaches  zero  when  the  number  of 
sides  approaches  infinity. 

(c)  Let  any  two  circles  be  taken,  and  let  the  radii  be  iJ,  r.  Let  AB  be  a 
side  of  a  regular  polygon  of  n  sides  inscribed  in  the  circle  having  centre  O 


¥iQ.  93. 


and  radius  jB,  and  let  ah  be  the  side  of  a  regular  polygon  of  n  sides  inscribed 
in  the  circle  having  centre  o  and  radius  r.  Let  P  denote  the  perimeter  of  the 
first  polygon,  p  that  of  the  second ;  let  C  denote  the  length  of  the  first  circle, 
and  c  that  of  the  second.    Then 

P=G-D,  p  =  c-d, 
where  D,  d  may  each  be  made  smaller  than  any  assignable  quantity  by  mak- 
ing the  number  of  sides,  ?i,  infinitely  great. 

The  polygons  are  similar,  since  they  are  regular  and  have  the  same  num- 
ber of  sides.     Hence,  by  geometry, 

P^OA^B, 
p      oa       r ' 

that  is,  ^-^  =  R. 

c  —  d       r 

.      From  this,  rC -rD-Bc-  Rd\ 

Whence,  rC  -  i2c  =  rD  -  i?d. 


APPENDIX. 


173 


Now,  let  n  become  infinitely  great.  Then  the  second  member  becomes 
smaller  than  any  assignable  quantity,  since  r,  J?,  each  remains  finite,  and 
d,  Z>,  each  approaches  zero.     Hence,  when  n  is  infinitely  great, 

rC  -Ec  =  0.  (1) 

rrom(l),  ^  =  ^,  and  1  =  1  (2) 

The  first  of  equations  (2)  may  be  expressed  in  words :  lengths  of  circles 
are  to  one  another  as  their  radii.  According  to  the  second  equation,  the 
length  of  the  first  circle  is  to  its  radius  as  the  length  of  the  second  circle  is  to 
its  radius.  But  these  are  any  two  circles.  Hence,  the  ratio  of  the  length  of 
a  circle  to  its  radius,  and,  consequently,  to  its  diameter,  is  constant.  The 
ratio  of  the  length  of  a  circle  to  its  diameter,  which  ratio  is  denoted  by  tt, 
will  now  be  approximately  determined.     [See  Arts.  9  (6),  (c),  72.] 

3.  The  formulas  used  in  this  determination  of  ir  are  deduced  in  problems 
Af  B,  that  follow  : 

A,  Given  the  radius  of  a  regular  inscribed  polygon,  to  compute  the  side 
of  a  similar  circumscribing  polygon. 

Let  AB  be  the  side  of  the  inscribed  polygon,  and 
OC  =  B,  the  radius  of  the  circle  ;  let  LM  be  a  side 
of  the  similar  circumscribing  polygon.    Let  LMhe 
obtained  by  producing  OA,   OB,  to  intersect  the 
tangent  drawn  at  C,  the  middle  point  of  the  arc  AB. 
The  triangles  LCO,  AEO,  are  similar.     Hence, 
LC  ^  PC 
AE      Oe' 

OCxAE     BxAE 


.'.  LC  = 


LM 


OE 

E  X  AB 

OE 


OE 


In  the  right-angled  triangle  OAE, 


OE 


Vo^^^aF=\ 


B2 


.'.  LM  = 


4 
2B  X  AB 

V4  B2  -  AB^ 


=  lV4i22 
2 


AB\ 


(1) 


J5.  Given  the  radius  and  the  side  of  a  regular  inscribed  polygon,  to  com- 
pute the  side  of  the  regular  inscribed  polygon  of  double  the  number  of  sides. 

In  Fig.  93,  let  AB  be  the  side  of  a  regular  inscribed  polygon  of  n  sides. 
Draw  AC ;  then  ^C  is  the  side  of  a  regular  inscribed  polygon  of  double  the 
number  of  sides,  namely,  2  n  sides.  It  is  required  to  express  AC  in  terms 
of  the  radius  B  and  the  side  AB.  Produce  CO  to  D  and  draw  DA.  The 
triangles  ACD^  AGE^  are  similar,  since  the  angles  DAC^  AEC,  are  equal, 


174 


PLANE  TRIGONOMETRY. 


both  being  right  angles,  and  the  angle  ACE  is  common  to  both  triangles. 
Hence, 

CD  :  AC  =  AG:  CE. 

.-.  AC^=  CD'  GE^  CDi^CO  -  E0)=1  R{B  -  E0)=  R{2  B  -2  EO). 


But 


EO  =Vo^^  -  ae'  =yR2  -  ^  =  I V4  i?2  -  AB". 


AC^  =  B(i2  B  --^4.  B^  -  AJ^). 
.  AC  =  \'i?C2 B-^4: B^  -  AB^) 


(2) 


4.  To  determine  approximately  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter.  If  the  radius  is  1,  the  length  of  the  circle  is  2  tt, 
and  the  length  of  the  semicircle  is  tt.  Hence  the  length  of  the  semi- 
perimeter  of  each  inscribed  and  circumscribing  regular  polygon,  is  an  approxi- 
mate value  of  TT,  and  approaches  nearer  and  nearer  to  tt,  the  greater  the 
number  of  sides  in  the  polygon.  The  side  of  the  inscribed  square  of  the 
circle  of  radius  1  is  V2  ;  its  semi-perimeter  is  2.8284271.  Successive  appli- 
cations of  (2),  Art.  3,  give  the  sides  of  the  inscribed  polygons  of  8, 16, 32,  — 
sides,  and  successive  applications  of  (1)  give  the  sides  of  the  similar  circum- 
scribing polygons.  The  successive  semi-perimeters  are  obtained  by  taking 
one-half  the  product  of  the  length  of  a  side  and  the  number  of  sides  in  the 
polygons.  The  results  of  the  computation  are  given  in  the  following  table. 
The  table  also  gives  the  results  when  the  initial  polygon  taken,  is  the  inscribed 
hexagon.     The  figures  in  bold  type  show  the  approximations. 

Lengths  of  semi-perimeters  of  regular  inscribed  and  circumscribing  poly- 
gons of  circle  of  radius  =  1. 


OF  Sides. 

Inscribed. 

Circumscribing. 

OF  Sides. 

Inscribed. 

CiRCUMSCKIBlNG. 

4 

2.8284271 

4.0000000 

6 

3 

3.4641016 

8 

3.0614675 

3.3137085 

12 

3.1058285 

3.2153903 

16 

3.1214452 

3.1825979 

24 

3.1326286 

3.1596599 

32 

3.1365485 

3.1517240 

48 

3.1393502 

3.1460862 

64 

3.1403312 

3.1441184 

96 

3.1410319 

3.1427146 

128 

3.1412773 

3.1422236 

192 

3.1414524 

3.1418730 

256 

3.1416138 

3.1417504 

384 

3.1415576 

3.1416627 

512 

3.1415729 

3.1416321 

768 

3.1415838 

3.1416101 

1024 

3.1415877 

3.1416025 

1536 

3.1416904 

3.1415970 

2048 

3.1415914 

3.1415951 

4096 

3.1415923 

3.1415933 

8192 

3.1415926 

3.1415928 

APPENDIX.  175 

5.  Area  of  a  circle.  Area  of  a  circular  sector.  The  area  of  a  circum- 
scribing polygon  of  a  circle  is  equal  to  one-half  the  product  of  the  lengths  of 
the  perimeter  and  the  radius.  When  the  number  of  sides  of  the  polygon 
increases  indefinitely,  the  perimeter  of  the  polygon  approaches  the  length 
of  the  circle  as  its  limit,  and  the  area  of  the  polygon  approaches  the  area 
contained  by  the  circle  as  its  limit.     Hence, 

area  of  circle  =  \  length  of  circle  x  length  of  radius ; 

i.e.  area  of  circle  =  lx2irBx  B  =  iriJ'^. 

Since  the  area  of  a  sector  of  a  circle  has  the  same  ratio  to  the  area  of  the 
circle  that  the  arc  of  the  sector  has  to  the  length  of  the  circle, 

area  of  circular  sector  =  \  length  of  arc  x  length  of  radius. 

Hence,  if  6  is  the  radian  measure  of  the  angle  of  the  sector, 

area  sector  =  \Bdx  B  =  fiJ^e. 

[For  example,  see  Art.  73,  Ex.  17.] 

6.  Historical  Note.  The  problem  to  find  a  square  whose  area  is  equal  to 
that  of  a  given  circle,  which  is  commonly  known  as  "squaring  the  circle," 
or  "the  quadrature  of  the  circle,"  has  long  been  of  interest  to  mathematicians 
and  others.  Since  the  area  of  a  circle  is  one-half  the  radius  by  the  length  of 
the  circle,  and  the  ratio  of  the  length  of  the  circle  to  the  diameter  is  a  con- 
stant, it  follows  that  "squaring  the  circle  "  comes  to  determining  this  ratio. 

The  ancient  peoples  used  3  as  the  value  of  tt  ;  see  1  Kings  vii.  23, 
2  Chron.  iv.  2.  Ahmes  used  3.1604;  Archimedes  showed,  by  the  method 
described  above,  and  by  successively  inscribing  and  circumscribing  regular 
polygons  of  6,  12,  24,  48,  96  sides,  that  -k  is  between  3ff  and  3f.  Ptolemy 
used  3^^,  and  Aryabhatta,  3.1416.  Adrian  of  Metz,  in  1527,  by  using  poly- 
gons up  to  1536  sides,  showed  that  the  ratio  is  between  f|^  and  |^|.  By 
taking  the  mean  of  the  numerators  for  a  new  numerator,  and  the  mean  of 
the  denominators  for  anew  denominator,  he  obtained  the  value  ff|,  which  is 
correct  to  six  places  of  decimals.  In  1579,  Vieta  by  using  polygons  of  32,316 
{i.e.  6  X  216)  sides,  got  the  value  of  tt  correctly  to  ten  places.  His  method  is 
not  the  same  as  that  of  Archimedes.  (Professor  Newcomb  has  remarked  that 
the  value  of  tt  to  ten  places  of  decimals  would  give  the  circumference  of  the 
earth  correctly  to  within  a  fraction  of  an  inch,  if  the  diameter  were  accurately 
known.)  In  1593,  Adrian  Roman  us  of  Louvain  computed  tr  to  15  places  of 
decimals  and  Ludolph  van  Ceulen  (d.  1610),  a  German  residing  m  Holland, 
calculated  it  to  35  places.  Hence  tt  is  often  referred  to  in  Germany  as  "  the 
Ludolphian  number." 

The  discoveries  of  trigonometric  series  (see  Note  A)  made  the  work  of  com- 
puters easier  and  more  mechanical.     In  1699,  Abraham  Sharp  (1651-1742), 


176  PLANE  TRIGONOMETRY, 

found  IT  to  71  places,  and  in  1706,  John  Machin,  died  1751,  professor  of 
astronomy  at  Gresham  College,  London,  extended  the  value  to  100  places. 
Fautet  de  Lagny  (1660-1734)  carried  it,  in  1719,  to  127  places ;  Baron  Georg 
Vega  (1756-1802),  in  1794,  to  136  places;  Z.  Dase  of  Vienna,  in  1844,  to 
200  places;  William  Rutherford  (1798  ?-1871),  Royal  Military  Academy, 
Woolwich,  in  1853,  to  440  places  ;  Richter,  in  1854,  to  500  places ;  and 
W.  Shanks,  in  1873,  to  707  places  of  decimals.  The  laborious  calculations 
of  the  "  TT-computers  "  have  neither  theoretical  nor  practical  value. 

About  1761  Lambert  showed  that  tt  is  incommensurable,  and  in  1882, 
F.  Lindemann,  in  Freiburg,  showed  that  it  is  transcendental,  that  is,  it  cannot 
be  a  root  of  any  algebraic  equation  with  integral  coefficients.  See  article, 
"  Squaring  the  circle,"  Ency.  Brit.,  9th  edition. 

N.B.  The  ratio  w  is  often  calculated  approximately  by  means  of  Gregory's 
series  (discovered  in  1670)  and  certain  identities,  namely, 

tan-ix  =  X  -^x^  +  ^x^  —  ^x"^  H toco,    ^  =  4tan-i^  —  tan-i^^, 

,  4 

-  =  tan-i  I  +  tan-i  I  +  tan-i  |,  -  =  4  tan-i  |  -  tan-i  ^  +  tan-i  ^^ 

in  which  the  principal  values  of  the  inverse  tangents  are  taken. 

The  student  is  advised  to  verify  these  identities,  and  to  find  an  approximate 
value  of  TT  by  means  of  Gregory's  series  ;  also  to  verify  the  remark  made 
above  concerning  the  circumference  of  the  earth.  Also  to  show  that  the 
method  of  Ahmes  (see  Note  A)  for  finding  the  area  of  a  circle  is  equivalent 
to  taking  IT  =  3.1604.... 


NOTE  D. 

DE  MOIVRE'S  THEOREM,  AND  OTHER  RESULTS  IN  ANALYTICAL 
TRIGONOMETRY. 

[In  what  follows  i  denotes  V— 1.] 

1.   De  Moivre's  Theorem.    For  all  values  of  n,  positive  and  negative, 
integral  and  fractional, 

(cos  6  +  1  sin  0)**  =  cos  nB  +  i  sin  wO. 

(a)   When  n  is  a  positive  integer. 
(cos  dy  +  i  sin  ^i)(cos  62  +  i  sin  ^2) 

=  COS  dx  cos  02  —  sin  di  sin  $2  +  i  (sin  di  cos  62  +  cos  di  sin  ^2) 
cos(^i  +  62)  +  i  sin(^i  +  ^3).  (1) 


APPENDIX,  111 

On  multiplying  each  member  of  (1)  by  cos  dz  +  i  sin  ^3,  there  is  obtained, 
(cos  di  +  i  sin  d{)  (cos  ^2  +  i  sin  ^2)  (cos  6^  +  i  sin  ^3) 

={cos(6'i  +  62)  +  i  sin(^i  +  ^2)}  (cos  dz  +  i  sin  ^3) 
={cos(^i  +  ^2)  cos  ^3  —  sin(^i  +  d^)  sin  ^3} 

+  i{sin(^i  +  62)  cos  dz  +  cos(^i  +  6^)  sin  ^3} 
=  cos  (^1  4-  ^2  +  ^3)  +  *■  sin(^i  +  ^2  +  ^3). 

In  a  similar  way,  the  product  of  four  or  more  such  factors  can  be  found. 
Thus,  for  n  factors, 

(cos  di  +  i  sin  ^1) (cos  62  +  i  sin ^2) •••(cos  dn  +  i sin  ^„) 

=  cos(^i  +  ^2  +  —  +  dn)  +  i  sin(^i  +  <?2  +  —  +  ^n)  (2) 

If  $1  =  02  =  •••  =  0n  =  01  then  (2)  becomes 

(cos  ^  +  i  sin  ^)«  =  cos  w0  +  i  sin  w^.  (3) 

(6)  When  n  is  a  negative  integer,  say,  —  m. 
(cos0  +  isin^)-"* 


(cos  d  +  i  sin  ^)"*     cos  md  +  i  sin  w^ 

1  cos  md  —  1  sin  md     cos  m^  —  i  sin 


cos  771^  +  i  sin  m^    cos  ?7i0  —  i  sin  m^     cos^  md  +  sin^  »h^ 
=  cos  md  —  i  sin  md  =  cos(—  m)^  +  i  sin(—  w)^.  (4) 

(c)  In  (3)  let  n9  =  (f>]  then  0  =  —^  and  (3)  becomes 

(0      .  .    0N"  .  . 

cos  -  + 1  sm  -     =  cos  0  +  I  sm  0. 
n  nj 

On  transposing  and  taking  the  nth  root  of  each  member  of  this  equation, 
there  is  obtained 

(cos  0  +  i  sin  0)"  =  cos  -  +  i  sin  -•  (6) 

n  n 

(The  second  member  of  (5)  is  one  of  the  n  roots  of  the  first  member.) 

n 

(d)  When  n  is  a  fraction,  — • 

p  1  1 

(cos  d  -\-i  sin  0)<'  =  [(cos  d  +  i  sin  ^)p]«  =  (cos;?^  +  i  sin pd)9 

=  cos^d  +  i  sin  ^  ^.  (6) 

(The  second  member  of  (6)  is  one  of  the  q  roots  of  the  first  member.) 


178  PLANE   TRIGONOMETRY. 

For  all  the  roots  of  the  first  members  of  (5),  (6),  see  one  of  the  works 
referred  to  in  Art.  98.  For  a  geometrical  representation  of  the  factors 
considered  above  and  the  results  (l)-(6),  and  for  a  proof  of  these  results, 
see  Hobson,  Flane  Trigonometry,  Chap.  Xlll. ;  Chrystal,  Algebra,  Part  I. , 
Chap.  XII. 

2.  Following  are  some  of  the  theorems  proved  in  analytical  trigonometry, 
which  will  be  met  by  those  who  read  only  a  little  farther  in  mathematics  : 

COSa;=l-  — +  ^^ to  00.  (1) 

sinx  =  a;-^  +  |i to  oo.  (2) 

01  O  I 

Expansions  (1)  and  (2)  were  first  shown  by  Newton  in  1669. 

If  e"  denote  the  series  1  +  x  +  —  +  ^  +  •••  to  oo,  (3) 

2  1      31 

pix  j_  p—ix       .  pix  p—ix 

then  cos  x  =  ^—Ll —  sin  x  = —  (4) 

2  2i  ^  ^ 

Formulas  (4)  were  first  given  by  Euler.  The  expansions  (1),  (2),  (3), 
are  also  derived  in  works  on  the  differential  calculus.  They  are  convergent 
for  all  finite  values  of  x.  Either  (1),  (2),  or  (3),  (4),  may  be  taken  as 
definitions  of  the  sine  and  cosine. 

Hyperbolic  functions.  The  hyperbolic  sine  and  cosine  of  x,  denoted  by 
sink  x,  cosh  x,  may  be  defined  in  either  one  of  the  following  ways,  namely, 


rr2       y4 

cosh  ic  =  l+  —  +  —  +...  to  00 
2!     41 

sinhx  =x-h  —  +  —  +  ...to  00 
31     61 


(5) 


and  coshx  =  ^^i^-^,  sinhic  =  ^^ — ^  (6) 

2  2 

A  geometrical  definition  may  also  be  given  to  the  hyperbolic  functions. 
In  this  definition,  they  are  related  to  the  hyperbola  in  a  manner  analogous 
to  a  way  in  which  the  trigonometric  (circular)  functions  are  related  to  a  circle. 
It  may  be  said  that  the  formulas  or  definitions  (l)-(C)  may  be  applied  to  all 
numbers  x,  real,  pure  imaginary,  or  complex;  i.e.  quantities  of  the  form 
a  +  6  V—  1.  (When  x  is  real  in  (l)-(4),  it  denotes  the  radian  measure 
of  the  angle.)     See  Chrystal,  Algebra,  Part  II.,  Chap.  XXIX.* 


APPENDIX.  179 


EXERCISES. 

1.  Substitute  1  for  x  in  the  series  (3),  and  thus  deduce  2.71828  as  an 
approximate  value  of  e. 

2.  (a)  Write  the  series  for  e»"*  and  e-^  by  substituting  ix  and  —  ix  for 
X  in  (3). 

(&)  Then  find  the  value  of  cos  x  in  (4),  and  compare  the  result  with  (1). 
(c)  Then  find  the  value  of  sin  x  in  (4),  and  compare  the  result  with  (2) . 

3.  Using  formulas  (4),  show  that  cos^aj  +  sin^aj  =  1. 

4.  Substitute  ix  for  x  in  (1)  and  (2),  and  compare  the  results  with  (5). 
Substitute  ix  for  x  in  (4),  and  compare  the  results  with  (6).  Each  substitu- 
tion shows  that  cos  {ix)  =  cosh  x,  and  sin  (ix)  =  i  sinh  x. 

6.  Show  by  means  of  the  formulas  (l)-(4)  that  the  cosine  of  an  angle 
of  magnitude  zero  is  unity,  and  that  the  sine  of  such  an  angle  is  zero. 

6.  By  means  of  (1),  (2),  find  approximate  values  of  sin  10°,  cos  10°, 
sin  15°,  cos  15°,  sin  20°,  cos  20°,  sin  30°,  cos  30°.  [First,  express  the  angles 
in  radian  measure.] 

*  Also  see  McMahon,  Hyperbolic  Functions,  (Merriam  and  Woodward, 
Higher  Mathematics,  Chap.  IV.,  pp.  107-168). 


QUESTIONS  AND    EXERCISES    FOR    PRACTICE 
AND   REVIEW. 


3><K< 


It  is  not  intended  that  all  these  exercises  be  worked  by  any- 
one person,  or  by  any  one  class.  It  is  advisable  to  consider  only 
a  few  of  them  on  the  completion  of  each  chapter,  and  to  use  them 
chiefly  in  the  general  reviews.  In  each  set  there  are  a  number  of 
direct  questions  on  the  principles  and  theorems  explained  in  the 
corresponding  chapter;  these  questions  will  enable  the  pupil  to 
examine  himself  concerning  his  knowledge  of  the  text.  Teachers 
will  often  do  well  to  take  examples  from  other  sources. 

CHAPTER   I. 

1.  (a)  Define  and  illustrate  logarithms^  characteristic,  mantissa, 
(b)  What  is  meant  by  the  base  of  a  system  of  logarithms?  (c)  What  is 
meant  by  a  system  of  logarithms  ?  (d)  Show  that  in  all  systems  log  1  —  0, 
and  that  the  logarithms  of  all  proper  fractions  are  negative. 

2.  What  are  the  advantages  gained  by  the  use  of  logarithms  calculated  to 
the  base  10  ?  Show  that  the  characteristic  of  any  logarithm  to  the  base  10 
may  be  found  by  inspection.  ^ 

3.  What  are  the  logarithms  to  base  3,  of  81,  ■^,  v^729  ? 

4.  Find,  by  using  logarithms,  the  values  of  the  following  quantities  : 
(a)  \/375,  VSTTS,  V3?75,  V:375,  V^0375;  (6)  </T2:5,  VlM,  </A2E, 
v'.0"l25,  ^:00l25;  (c)  \/784,  ^/93,  Vssi;  (d)  (19)^  (212)*,  (31.7)^; 
(e)    4-7,  5"^,  (67) "i 

6.  The  following  calculations  are  to  be  made  at  first  without  the  use  of 
logarithmic  tables.  The  results  may  then  be  checked  by  comparing  them 
with  the  values  obtained  by  means  of  the  tables,  (a)  Given  log  3  =  .477121, 
find  log  {(2.7)3  X  (81)^  -(90)^}.  (&)  Given  log  5  =  .69897,  find  log  200, 
log  .025,  log  vfes.  (c)  Given  log 2  =  .30103,  find  the  logarithms  of  5,  i|^, 
VaM.     (d)    Given  log  2  =  .3010   and  log  3  =  .4771,   find  log  ^j,  log  .25, 

181 


182  PLANE   TRIGONOMETUY. 

log  16.2,  log  Vf.  (e)  Given  log  2  =  .3010,  log  3  =  .4771,  log  5  =  .6990, 
log  7  =  .8451,  find  the  logarithms  of  f^,  175,  .0054,  (12)^,  v^35.  (/)  Given 
logs  =  .903,  log9  =  .954,  find  the  logarithms  of  2,  3,  12,  500,  .075. 

6.  Find  by  logarithms  the  values  of  the  following  quantities  : 

{a)    372.48  x  (i|9«^    (.006)5  ^  125;  (&)    3487  x  (.00345)^  -(-  83)^; 

r.\     ^/3       y-V       ^^N  / 345.4  X  958.3 U  /417.9  x  813.1  \i       ,.     /oq,nI 

(c)  Vf-^V,^,;  (^)U3.4x  317.9)-  (964.7x313.2)  -  ^'^  ^''^^^^ 
(411)^- (7.93)3-5. 

7.  Find  an  approximate  value  of  x  in  each  of  the  following  equations : 
{a)  x2  =  237,  (6)  x^  =  17,  (c)  x-^  =  17,  (d)  x-^  =  41,  (e)  2^  =  9, 
(/)3-=197,  (^)3-  =  32,  (/i)(25)3-2x  =  2-+^  (i)  log(a:2)  +  log(2x)  + 1  =0. 

8.  If  the  logarithm  of  27  is  —  f ,  what  is  the  base  ? 

CHAPTER   II. 

1.  If  on  a  map  a  square  inch  represents  10  acres,  how  many  yards  are 
represented  by  the  diagonal  of  a  square  inch  ? 

2.  Explain  the  English  and  French  methods  of  measuring  angles,  and 
show  how,  when  the  measure  of  an  angle  according  to  either  method  is 
known,  its  measure  according  to  the  other  may  be  found.  Express  100°  in 
grades.     (See  Note  2,  Art.  11.) 

3.  If  A  is  an  acute  angle,  show  that  tan  J.  is  greater  than  sin  A. 

4.  By  aid  of  an  equilateral  triangle  find  the  numerical  values  of  the  six 
trigonometric  ratios  of  60°  and  30°.  Find  the  numerical  values  of  the  ratios 
of  45°. 

5.  Show  that  («)  a/^!"  tTo  ~4^  =  sec  45°-  tan  45°, 

>'sin45°  +  sin30°  * 

,^.  1  +  cot  60°  _  /  1  +  cos  30° 


l-cot60°       U-cos30° 
(c)  tan2  60°  -  2  tan2  45°=  cot2  30°  -  2  sin2  30°  -  f  cosec2  45°. 

6.  The  sine  of  an  angle  defined  as  a  ratio  being  less  than  unity,  explain 
why  the  tabular  logarithms  of  the  sines  of  angles  are  expressed  with  whole 
numbers  as  characteristics.  Given  log  tan  18°  =  9.51178,  show  what  the 
tabular  logarithm  of  cot  18°  must  be. 

7.  (a)  Given  log  2  =  .30103,  log  3  =  .47712,  find  log  sin  60°  and  log  tan  30°. 
(6)  Given  log  5  =  .69897,  find  the  logarithmic  sine  of  30°,  and  the  logarithmic 
cosine  of  45°. 

8.  Compute  the  trigonometric  ratios  of  yl  in  a  right  triangle  ABC{C=  90°), 
when  b  =  ^c. 

9.  Construct  the  following  right-angled  triangles :  (a)  ABC,  in  whicli, 
C  =  90°,  c  =  5,  cot  J.  =  I ;  (&)  when  one  of  the  legs  is  3,  and  the  sine  of  the 
adjacent  acute  angle  is  f ;  (c)  hypotenuse  4,  and  sine  of  one  of  the  acute 


qu:estions  and  exercises,  183 

angles  f ;    (d)  0  =  90°,  sin  ^  =  f ,  6  =  7;    (e)  C  =  90°,  cosec  ^  =  f ,  b  =  10  ; 
write  the  values  of  sin  A,  cos  ^,  tan  A  ;  (/)  (7  =  90°,  cos^  =  |,  a  =  9. 

10.  In  the  triangle  ABC,  C  =  90°,  tan  B  =  if.     If  AB  =  510  ft.,  find  AC. 

11.  In  ABC,  C=  90°,  J5(7  =  10  ft.,  tan  5  =  1.05  ;  find  the  other  sides. 

12.  The  string  of  a  kite  is  250  ft.  in  length.  How  high  is  the  kite  above 
the  ground  when  the  string,  supposed  stretched  quite  tight,  makes  with  the 
ground  an  angle  whose  tangent  is  ^^  ? 

13.  ABC  is  an  isosceles  triangle,  right-angled  at  (7;  Z>  is  the  middle  point 
of  AC.  Prove  that  DB  divides  the  angle  B  into  two  parts  whose  cotangents 
are  as  2  : 3. 

14.  (a)  Given  L.  cos  20°  =  9.97  and  L.  cot  20°  =  10.44 ;  find  each 
of  the  other  logarithmic  ratios  of  20°.  (b)  Given  L.  sin  40°  =  9.808, 
L.  tan  40°  =  9.924  ;  find  log  cot  40°,  log  cos  40°,  log  sec  40°,  log  cosec  40°. 

2  Vab  sin  — 

15.  If  tan  e  = ;— ?- ,  find  d  when  a  =  5,  5  =  2,  C=  120°. 

a  ~  b 


16.  Calculate  sin3  23°  x  V27.268  -4-  2  cos2  48°. 

17.  Find  x  in  the  equations :  (a)  x  sin  74°  =  235  tan  37°  cos  63°, 

(&)  x^  cos  39°  =  47.5  sin2  46°  sec2  64°. 

18.  Solve  (sin  8°  +  cos  8°)2^  =  2  sin  16°  (tan  32°)^ 

CHAPTER  III. 

1.  State  what  parts  of  a  right  plane  triangle  must  be  given  that  it  may 
be  constructed,  and  show  how  a  right  triangle  may  be  solved,  in  each  of 
the  four  possible  cases. 

2.  Derive  the  formulas  for  computing  B,  a,  and  c  of  a  right  triangle  when 
C  =  90°,  and  A  and  b  are  given.  Also  find  a  formula  that  shall  include  only 
the  required  parts. 

3.  In  ABC,  C  =  90°,  &  =  22  ft.,  and  sin  A  =  .42.  Find  a,  c,  sin  B,  and 
the  area. 

4.  In  ABC,  C  =  90°,  cosA  =  ^,  c  =  40  ft.  Find  the  values  of  cos  B, 
cot  B,  a,  b,  and  the  area. 

5.  Solve  the  following  right-angled  triangles  by  (1)  making  an  ofE-hana 
estimate,  (2)  measuring  on  a  drawing  made  to  scale,  (3)  computing  without 
logarithms  (four-place  tables),  (4)  computing  with  logarithms.  Check  the 
results  by  computation.  If  a  solution  is  impossible,  explain  why  it  is  so. 
(Each  triangle  is  denoted  by  ABC,  and  (7  =  90°.)  (i.)  a  =  45,  &  =  62  ; 
(ii.)  a  =  685,  5  =  34°  47' 25";  (iii.)  c  =  560,  a  =  310  ;  (iv.)  c  =  327,6  =  450; 
(V. )  c  =  520,  A  =  36°  40'  20"  ;  (vi. )  6  =  720,  B  =  61°  24'  30"  ;  (vii.)  c  =  425, 
B  =  32°  45'  35"  ;  (viii.)  a  =  11524,  6  =  35976  ;  (ix.)  a  =  67213,  6  =  75324  ; 
(X.)  c  =  35421,  6  =  23462. 


184  PLANE  TBIGONOMETBY. 

6.  Two  sides  of  a  triangle  are  as  5  :  9,  and  the  included  angle  is  a  right 
angle.     Find  the  other  angles. 

7.  Find  the  acute  angles  of  a  right-angled  triangle  whose  hypotenuse  is 
six  times  as  long  as  the  perpendicular  let  fall  upon  it  from  the  opposite 
angle. 

CHAPTER  IV. 

1.  (a)  Derive  the  formula  for  the  area  of  a  right  triangle  in  terms  of 
(i.)  an  angle  and  its  opposite  side,  (ii.)  an  angle  and  its  adjacent  side. 
(&)  One  side  of  a  triangle  is  seven  times  another,  and  the  included  angle  is  a 
right  angle.     Find  the  other  angles. 

2.  Show  how  an  isosceles  triangle  may  be  divided  into  right  triangles,  and 
how  it  may  be  solved  by  aid  of  these  right  triangles  when  the  following 
elements  are  given :  (a)  Base  and  vertical  angle,  (&)  base  and  side,  (c)  side 
and  vertical  angle,  (d)  base  and  perpendicular  from  vertex  on  the  base. 
Discuss  any  other  possible  cases. 

3.  Solve  the  isosceles  triangles  (a)  whose  base  is  126  ft.,  and  vertical 
angle  is  127° ;  (6)  whose  base  and  perpendicular  on  it  from  the  vertex  are 
each  721.34  yd. 

4.  (a)  Find  the  area  of  a  regular  octagon  the  side  of  which  is  26  yd. 
(&)  Find  the  side  of  a  regular  pentagon  inscribed  in  a  circle  whose  radius  is 
43  ft.  (e)  If  a  regular  pentagon  and  a  regular  decagon  have  the  same 
perimeter,  prove  that  their  areas  are  as  2  :  Vb. 

v/  5.  (a)  At  120  ft.  distance,  and  on  a  level  with  the  foot  of  a  steeple,  the 
angle  of  elevation  of  the  top  is  62°  27' ;  find  the  height,  (h)  From  a  cliff 
330  ft.  high  the  angle  of  depression  of  a  boat  at  sea  is  40°  35'  25"  ;  how  far 
is  the  boat  from  the  foot  of  the  cliff  ? 

6.  When  the  altitude  of  the  sun  is  30°  the  length  of  the  shadow  cast  by 
Bunker  Hill  monument  is  381  ft.     What  is  the  height  of  the  monument  ? 

7.  The  angles  of  depression  from  the  top  of  a  tower  48.6  ft.  high  to  two 
points,  on  a  level  with  its  base  and  in  line  with  the  tower,  are  45°  and  30° 
respectively.  Find  the  distances  of  each  point  from  the  other  and  from  the 
top  of  the  tower. 

8.  A  pole  40  ft.  high  is  erected  at  the  intersection  of  the  diagonals  of  a 
square  courtyard.  When  the  sun's  altitude  is  43°  40',  the  shadow  just 
reaches  a  corner  of  the  yard.     Find  the  length  of  the  side  of  the  square. 

9.  (a)  When  the  altitude  of  the  sun  was  67°  30'  45"  the  length  of  the 
shadow  of  a  perpendicular  pole  was  73.4  ft.  Find  the  length  of  the  shadow 
when  the  sun's  altitude  is  35°.  (&)  The  shadow  of  a  tower  is  observed  to 
be  half  the  known  length  of  the  tower,  and  some  time  after  to  be  equal  to  the 
full  length.     How  much  will  the  sun  have  gone  down  in  the  interval  ? 


qu:estions  and  exercises,  185 

10.  A  flagstaff  which  leans  to  the  east  is  found  to  cast  shadows  of  198  ft. 
and  202  ft.,  when  the  sun  is  due  east  and  west  respectively,  and  his  altitude 
is  7°.     Find  the  length  of  the  flagstaff  and  its  inclination  to  the  vertical. 

11.  What  angle  will  a  flagstaff  24  ft.  high,  on  the  top  of  a  tower  200  ft. 
high,  subtend  to  an  observer  on  the  same  level  with  the  foot  of  the  base,  and 
100  yds.  distant  from  it  ? 

12.  Looking  out  of  a  window  with  his  eye  at  the  height  of  15  ft.  above 
the  roadway,  an  observer  finds  that  the  angle  of  elevation  of  the  top  of  a 
telegraph  post  is  IT*^  18'  o5",  and  that  the  angle  of  depression  of  the  foot  of 
the  post  is  8°  32'  15".  Calculate  the  height  of  the  telegraph  post  and  its 
distance  from  the  observer. 

13.  A  man  in  a  balloon,  when  it  is  one  mile  high,  finds  the  angle  of 
depression  of  an  object  on  the  level  ground  to  be  35°  20',  then  after  ascending 
vertically  and  uniformly  for  20  min.,  he  finds  the  angle  of  depression  of 
the  same  object  to  be  65°  40'.  Find  the  rate  of  ascent  of  the  balloon  in  miles 
per  hour. 

14.  A  man  observes  the  elevation  of  a  mountain  top  to  be  15°,  and  after 
walking  3  mi.  directly  toward  it  on  level  ground,  the  elevation  is  18°.  Find 
his  distance  from  the  mountain. 

15.  From  a  boat  the  angle  of  elevation  of  the  highest  and  lowest  points  of 
a  flagstaff,  30  ft.  high,  on  the  edge  of  a  cliff  are  observed  to  be  46°  12'  and 
44°  13'.     Determine  the  height  of  the  cliff  and  its  distance. 

16.  The  angles  of  elevation  of  the  top  of  a  tower,  observed  at  two  points 
in  the  horizontal  plane  through  the  base  of  the  tower,  are  tan-i  |  and  tan-i  _5^  j 
the  points  of  observation  are  240  ft.  apart,  and  lie  in  a  direct  line  from  the 
base.     Find  the  height  of  the  tower. 

17.  A  person  standing  due  south  of  a  lighthouse  observes  that  his  shadow 
cast  by  the  light  at  the  top  is  24  ft.  long ;  on  walking  100  yd.  due  east  he 
finds  his  shadow  to  be  30  ft.  Supposing  him  to  be  6  ft.  high,  find  the 
height  of  the  light  from  the  ground. 

18.  An  observer  is  384  yd.  due  south  of  a  point  from  which  a  balloon 
ascended ;  he  measures  a  horizontal  base  due  east,  and  at  the  other  extremity 
finds  the  angle  of  elevation  to  be  60°  15'.     Find  the  height  of  the  balloon. 

19.  A  surveyor  starts  from  A  and  runs  766  yd.  due  east  to  J5,  thence 
622  yd.  N.  20°  30'  E.  to  O,  thence  850  yd.  N.  41°  45'  W.  to  D,  thence  S. 
42°  36'  W.  to  E.  Find  the  distance  and  bearing  of  A  from  E,  and  determine 
the  area  of  the  field  ABODE. 

20.  A  surveyor  runs  253  yd.  N.E.  by  E.,  thence  N.  by  E.  212  yd.,  thence 
W.N.W.  156  yd.,  thence  S.W.  by  S.  210  yd.,  thence  to  the  starting- 
point.  Find  the  bearing  and  distance  of  the  starting-point  from  the  last 
station,  and  determine  the  area  of  the  field  which  the  surveyor  has  gone 
around. 


186  PLANE  TRIGONOMETRY. 


CHAPTER   V. 

1.  Define  and  illustrate  angle,  negative  angle,  complement  of  an  angle, 
supplement  of  an  angle,  quadrant,  angle  in  the  third  quadrant. 

2.  Define  and  illustrate  the  six  trigonometric  ratios.  Find  the  greatest 
and  least  values  that  each  of  them  can  have.  Arrange  in  tabular  form  the 
algebraic  signs  of  the  trigonometric  ratios  of  an  angle  in  each  quadrant. 

3.  Explain  how  the  trigonometric  ratios  of  an  angle  of  any  magnitude, 
positive  or  negative,  can  be  found,  {a)  by  means  of  tables  which  give  these 
ratios  for  angles  up  to  90°  only,  (&)  by  means  of  tables  which  give  these 
ratios  for  angles  up  to  45°  only. 

4.  Prove  that  if  two  angles  have  the  same  sine,  and  also  any  of  the  other 
five  trigonometric  ratios  (with  one  exception)  the  same,  they  will  differ  by  a 
multiple  of  360°. 

5.  State  and  prove  the  chief  relations  which  exist  between  the  trigono- 
metric ratios  of  any  angle  A. 

6.  Express  the  trigonometric  ratios  of  90°  —  A,  90°  +  A,  180°  —  A, 
180°+^,  270°-^,  270°  +  ^,  360° -yl,  -A,  in  terms  of  the  trigono- 
metric ratios  of  A. 

t.  Name  three  pairs  of  trigonometric  ratios  such  that  the  product  of  each 
pair  shall  equal  1  ;  one  pair,  the  sum  of  whose  squares  shall  equal  1  ;  two 
pairs,  the  difference  of  whose  squares  shall  equal  1. 

8.  Compare  the  trigonometric  ratios  of  any  angle  (a)  with  those  of  its 
complement,  (6)  with  those  of  its  supplement. 

9.  Prove  that  sin  ^  =  cos  ^  tan  A  ;  sec^  ^  =  1  +  tan^  A  ; 
cot  A  =  cosec  A  cos  A;  sin2  ^  _|.  cos^  A  =  l;  sin  ^  =  tan  ^ :  VT+  tan^  e ; 
cos  X  =  Vcosec^  x  —  l:  cosec  x. 

10.  (a)  Express  the  following  trigonometric  ratios  in  terms  of  trigono- 
metric ratios  of  positive  angles  not  greater  than  45°  :  sin  237°,  cos  (—  410°), 
tan  2000°,  cot  (-137°),  sec  445°,  cosec (- 650°),  sin  185°,  tan 267°,  sec  345°, 
cos  87°,  cot  (  —  19°) ;  (6)  by  means  of  the  tables  give  the  numerical  values  of 
these  ratios. 

11.  Find,  without  the  use  of  trigonometric  tables,  the  numerical  values  of 
cos  1410°,  tan  (-1260°),  cosec  (- 1710°),  tan 225°,  cot  1035°,  cosec  210°, 
cos  1600°,  sin  1665°,  tan  (-1665°),  all  the  trigonometric  ratios  of  -1125° 
and  930°. 

12.  Construct  the  angles :  (a)  whose  secant  is  3,  (6)  whose  tangent  is 
V2  4-  1,  (c)  whose  cotangent  is  |.     Find  the  other  ratios  of  these  angles. 

13.  (a)  Find  sin  A,  cot  A,  when  cos  ^  =  -  j%,  and  A  <  180°.  (6)  Find 
the  other  ratios  of  A  and  x  when  cot  A  =  -^  and  cos  x  =  —  |.  (c)  Find  the 
other  ratios  of  A  when  cos  ^  =  —  ^,  and  A  lies  between  540°  and  630°. 


QUESTIONS  AND  EXEttCISES.  187    ^_^ 

((2)  Find  the  trigonometric  ratios  of  180°  +  6  and  270°  —  ^,  given  tan  6  =  \. 
(e)  Given  sec  a;  =  —  |,  and  x  in  the  third  quadrant ;   find  the  value  of 
sin  X  +  tan  x_ 
cos  X  +  cot  X 

14.  Do  Ex.  9,  Art.  18,  J.  being  any  angle.  Explain  the  ambiguities  in 
the  algebraic  signs.  If  A  is  an  angle  in  the  third  quadrant,  express  cos  A, 
tan  A,  cot  A,  sec  J.,  cosec  A  in  terms  of  sin  A. 

15.  (a)  If  sec  A  =  n  tan  A,  find  the  other  ratios  of  A.  (6)  If  2  sec  0  = 
tan  a  +  cot  a,  find  tan  0  and  cosec  d.     (c)  Solve  x^  cot  108°  =  128°  sin  72°  cos  18°. 

16.  Prove  the  identities  :  sin^  6  +  cos^  6  =  (sin  d  +  cos  6)  (1  —  sin  6  cos  6); 
cos-*  A  —  sin*  ^  =  1  —  2  sin^  A  ;  sin  x  (cot  x  +  2)  (2  cot  ic  +  1)  =  2  cosec  x  + 
5  cos  X  ;    sec2  5  —  cos^  B  =  cos^  i^  tan'^  B  +  sin^  J5  sec^  B  ;   cos^^  ^  +  sin^  A  = 

1  —  3  cos2  A  sin2  ^  ;  cos^  aj  +  2  cos*  x  sin^  x  +  cos^  x  sin*  x  +  sin^  x  =  1. 

17.  (a)  Find  the  value  of  x  not  greater  than  two  right  angles  which  will 
satisfy  the  equation  4\/3  cot  x=7  cosec  x— 4  sin  x.  (&)  Likewise,  in  the  case 
of  the  equation  sinx+cosxcotx=2.  (c)  Likewise,  in  tan*x— 4tan2x+3=0. 
(d)  If  1  +  sin2  ^  =  3  sin  6  cos  6,  find  tan  6.  (e)  Find  the  least  positive  value 
of  A  that  satisfies  the  equation  2\/3  cos^^  =  sin  J..  (/)  Find  all  the 
angles  between  0°  and  500°  which  satisfy  the  equation  4  sin^  ^  =  3.     (g)  If 

2  cos  J.  +  sec  ^  =  3,  what  is  the  value  of  ^  ?     (h)  Find  A  when  tan^  A  + 
cosec2  ^  =  3. 

CHAPTER  VI. 

1.  (a)  Write  the  values  of  cos(^+J5),  cos(A  —  B),  sm(A  +  B)^ 
sin(^  —  J5),  tan  (J.  +  B),  tan  (A  —  B)  in  terms  of  the  trigonometric  ratios 
of  A  and  B.     (6)  Deduce  these  values,     (c)  Express  them  in  words. 

2.  (a)  Express  in  terms  of  the  trigonometric  ratios  of  A  each  of  the 
following:  sin  2  J.,  cos  2^  (three  different  forms),  tan  2  J.,  cot  2  A 
(6)  Derive  these  expressions. 

3.  (a)  Show    that    sin  ^  +  sin  5  =  2  sin  A±^  cos  ^  ~  ^.     (&)    Show  ^ 

2  2  ^t~Z 

that  cos  2  ^  +  cos  2  5  =  2  cos  {A  +  B)  cos  (^A  —  B).     (c)   State   and  derive         ^ 

an  equivalent  expression  for  the  difference  of  two  sines  ;  (d)  for  the  differ- 
ence of  two  cosines. 

4.  Show  how  to   find   cos^J.  when  cos  J.    is    known.      Explain    the 

ambiguity  in  the  result.     Determine  the  sign  of  the  result  when  A  is  an 

angle  in  the  third  quadrant.     Find  the  cosine  of  112°  30'.     [From  cos  225°.] 
*  .         

5.  Prove  2  cos  ~  =  —  Vl  +  sin^  —  Vl  —  sin  J.    if    J.    is  between    270° 

and  360°.  ^ 

6.  Derive  an  expression  for  each  of  the  following  :  sin  3  ^  in  terms  of 
sin  A^  cos  3  A  in  terms  of  cos  A^  tan  3  J.  in  terms  of  tan  A.  [Suggestion  : 
3^=2^  +  ^.     See  Art.  93.] 


188  PLANE  TBIGONOMETRY. 

7.  (a)  If  tan  A=  and  tan .5=  ,  prove  that  tan (^-5)  =  .375. 

4-V3  4+\/3 

1  3 

(h)  If  sin  A  =  — ■::z  and  cos  B  =  -,  find  the  value  of  tan(^  +  B).    (c)  Find 

VlO  5 

tan  (J.  +  B),  given  that  sin  A  =  —,  sin  5  =  — . 

17  13 

8.  (a)  If  tan^  =  ^,  show  that  sin^=-l^o'  sin  2  J.  =  i«^<«i::i^. 
(6)  If  tan  A  =  K  prove  that  J«^  +  a/^^  =  J-^2^. 

9.  (a)  Find  sin  45°,  and  thence  deduce  the  ratios  of  22°  30'.  (6)  Prove 
that  tan  67°  30'  =  1  +  V2.  (c)  Deduce  the  ratios  of  67°  30',  (i.)  from  ratios 
of  45°,  22°  30',  (ii.)  from  ratios  of  135°. 

10.  (a)  Given  sin  30°  =  ^  and  cos45°  =  |\/2;  find  sin  15°,  cos  75°. 
(6)  Given  sin  30°  =  | ;  find  the  numerical  values  of  the  other  ratios  of  30°  ; 
thence  derive  the  ratios  of  15°,  thence  derive  the  ratios  of  75°,  105°,  165°, 
195°.  (c)  Prove  the  following  :  tan  15°  +  tan  75°  =  4,  cos  15°  .  cos  75°  =  .25, 
sin  105°+ cos  105°= cos  45°,  tan  15° (tan  60°- tan  30°) = tan  60°+ tan  30°-2. 

11.  (a)  Express  sin8^  +  sin2J.  as  a  product.  (6)  Express  as  a  sum 
or  difference  :  (i.)  2cos^cosJ5,  (ii. )  2  sin  50°  cos  20°.  (c)  Prove  without 
using  tables  that  (i.)  sin  70°  -  sin  10°  =  cos  40°,  (ii.)  cos  20°  +  cos  100°  + 
cos  140°  =  0.    Verify  by  the  tables. 

12.  Show  that:  (1)  cot^  cot  5cos(^+B)=cos  ^  cos  5(cot^cotS--l); 
(2)  cos  {A  +  B)  cos  A  +  sin  (A  +  B)  sin  ^  =  cos  J5  ;  (3)  cos  ^  -  sin  ^  = 
V2  cos(A  +  45°) ;  (4)  2  cos^  x-2  sin^  x  =  cos  2  x(l  +  cos2  2  x)  ;  (5)  cos2  A  + 
siii2  ^  cos  2  5  =  cos2  B  +  sin^  ^  cos  2  ^  ;  (6)  cos2  ^  -  cos  ^  cos  (60^  +  A)  + 
sin2  (30°  -  ^)  =  .75  ;  (7)  tan  ^  ^  =  sin  ^  :  1  +  cos  6  ;  (8)  cos  (135°  +  A)  + 
sin(135°  -  J)  =  0  ;    (9)  cosec  2  ^  +  cot  2  ^  =  cot  ^. 

13.  Prove      that:     (1)     sin  a;  +  sin  y ^_         .  .  .         ^2)     tan^  = 
^^     cos X- cosy                ^^        ^^'       ^^  2 

'2sin^-sin2^  ^^^  ^^^  _  8  cot  ^ 

2  sm  ^  +  sm  2  ^  ^  ^  \         «       y  v  j      ^^^^  ^  -  3 

(4)  gg^+^^°)=sec2^-tan2^-    (5^  cos2  Jg-cos2  vl^sin  2^-sin  2  g_ 
cos(^-45°)  '    ^  ^  sin2^+sin2^    cos2^  +  cos2i? 

tan(^-^);    (6)  sec2^-itan2^sin2^=-^"iiA±iH^. 

cot2  A  —  tan2  A 

a 

14.  (a)  Find  values  of  d  not  greater  than  180°,  which  satisfy  cot  ^=tan  -. 

(6)  Give  all  the  positive  angles  less  than  S60°,  which  satisfy  the  equation 
sin  2  ^  =  V3  cos  2  A. 

16.  Show  that  the  value  of  sin(n  +  l)5sin(n-l)B+cos(«  +  l)J5cos(n-l).B 
is  independent  of  n. 


4 


QViJSTlONS  AND  EXERCISES.  189   ^ 

16.  The  cosines  of  two  angles  of  a  triangle  ABC  are  |  and  i|,  respec- 
tively ;  find  all  the  trigonometric  ratios  of  the  third  angle  without  using 
tables.     Verify  the  results  by  means  of  the  tables. 

17.  Two  towers  whose  heights  respectively  are  180  and  80  ft.,  stand  on  a 
horizontal  plane  ;  from  the  foot  of  each  tower  the  angle  of  elevation  of  the 
other  is  taken,  and  one  angle  is  found  to  be  double  the  other  ;  prove  that  the 
horizontal  distance  between  the  towers  is  240  ft.,  and  show  that  the  sine  of 
the  greater  angle  of  elevation  is  .6. 

CHAPTER   VII. 

1.  In  a  triangle  ABC,  show  that  (1)  sin(^  +  B)=  sin  C,  (2)  cos(^  +  B) 

=  -cosO,  (3)sin^-±^  =  cos-^,   (4)  cos  "^  +  ^  =  sin  ^. 

2  2  2  '2 

2.  (a)  State  and  prove  the  Law  of  Sines  for  the  plane  triangle.  (6)  State 
and  prove  the  Law  of  Cosines  for  the  plane  triangle,  (c)  If  the  sines  of  the 
angles  of  a  triangle  are  in  the  ratios  of  13  :  14  :  15,  prove  that  the-  cosines  are 
in  the  ratios  of  39  :  33  :  25. 

3.  (a)  Prove  that  in  ABC,  6  +  c  :  6  -  c  =  tan  ^  (5  +  (7)  :  tan  |(i?  -  C) 
=  cot  \  ^  :tan  1{B—C).  (h)  Write  and  derive  the  expressions  for  the  cosine 
of  an  angle  of  a  triangle,  and  the  cosine  and  the  sine  of  half  that  angle,  in 
terms  of  the  sides  of  the  triangle,  (c)  In  the  triangle  ABC  derive  the  for- 
mulas expressing  tan^^,  tan  ^  i?,  tan  1(7,  in  terms  of  a,  6,  c.     {d)  Prove 

that  in  any  triangle  ABC,  sin  ^  =  —  \/s(s  —  a){s  —  b)(s  —  c). 

be 

4.  (a)  Show  how  to  solve  a  triangle  when  the  three  sides  are  given, 
(i.)  without  logarithms,  (ii.)  with  logarithms.  Derive  all  the  formulas 
necessary.  (&)  Do  the  same  when  two  sides  and  their  included  angle  are 
given,     (c)  Do  the  same  when  two  angles  and  a  side  are  given. 

5.  (a)  Explain  carefully,  and  illustrate  by  figures,  the  case  in  which  the 
solution  of  a  triangle  is  ambiguous,  {b)  Write  formulas  for  a  complete 
solution  and  check,  of  a  triangle,  when  two  sides  and  an  angle  opposite  to 
one  of  them  are  given.  How  many  solutions  are  there  ?  Discuss  fully  all 
cases  that  may  arise,  (c)  Given  the  angle  A,  and  the  sides  a  and  &  of  a 
triangle  ABC,  determine  whether  there  will  be  one  solution,  two  solutions, 
or  no  solution,  in  each  of  the  following  cases  :  (i.)  A  <  90°,  a  >  6,  (ii. )  -4  <  90°, 
a  =  b,  (iii.)  A  <  90°,  a  <  6,  (iv.)  A  >  90°,  «  >  &,  (v.)  4  >  90°,  a  =  b. 

6.  Show  by  the  trigonometric  fornuilas  that  the  angles  of  a  triangle  can 
be  found  when  the  ratios  of  the  three  sides  are  given.  Give  the  geometrical 
explanation. 

7.  Show  by  the  trigonometric  formulas  that  the  other  two  angles  of  a  tri- 
angle can  be  found  when  the  third  angle,  and  the  ratio  of  the  sides  contain- 
ing it,  are  known.     Give  the  geometrical  explanation. 


190  PLANE  TRIGONOMETRY, 

8.  Assuming  the  law  of  sines  for  a  plane  triangle,  prove  that  a  -{■  b  :c  = 
cos  l(A-  B):  sin  ^  O,  and  a  -  6  :  c  =  sin  ^  (^  -  £)  :  cos  ^  C. 

9.  (a)  UA:B:C  =  2:S:4,  prove  2  cos  ^  =  ^^^-     (&)  If  2  a  =  ft  +  c, 

B         C      1  2  0 

prove  tan  —  tan  —  =  -.    (c)  If  a,  b,  c,  the  sides  of  a  triangle,  be  in  arithmetical 

AC  B 

progression,  prove  that  2  sin  —  sin  —  =  sin—.     [Suggestion  :  Put  2  6  =  a  +  c.] 

10.  [In  each  of  the  examples  in  Ex.  10,  J.,  J5,  C,  denote  the  angles,  a,  6,  c, 
the  sides  of  the  triangle.]  Solve  the  following  triangles  (1)  by  making  an 
estimate,  (2)  by  the  method  of  construction,  (3)  by  computation,  without 
using  the  logarithms  of  the  trigonometric  ratios  (four-place  tables),  (4)  by 
computation.  Using  logarithms  (five-place  tables),  (5)  by  dividing  some  of 
the  oblique  triangles  into  right-angled  triangles.  Check  the  results  by  com- 
putation. When  a  solution  is  impossible,  or  ambiguous,  explain  why  it  is  so. 
(I)  a  =  753,  6  =  621,  c  =  937  ;  (2)  a  =  9,  &  =  17,  c=14;  (3)  a  =  1236.5, 
b  =  1674.8,  c  =  2532.7  ;  (4)  a  =  30,  &  =  42,  c  =  36  ;  (5)  a  =  621,  b  =  237, 
c  =  325  ;  (6)  a  =  1237,  b  =  1014,  A  =  39°  42' ;  (7)  a  =  1114,  b  =  1345, 
A  =  46°  54'  20"  ;  (8)  c  =  832,  b  =  694,  B  =  54°  47'  30"  ;  (d)  a  =  1020,  b  =  240, 
B  =  70°  25' ;  (10)  c  =  794,  b  =  832,  B  =  65°  30'  20"  ;  (11)  c  =  230,  a  =  950, 
C  =  63°  47' ;  (12)  a  =  237,  c  =  452,  C  =  37°  49' ;  (13)  a  =  420,  c  =  337,  C  = 
42°  46' ;  (14)  a  =  452,  b  =  624,  C  =  37°  23' ;  (15)  a  =  1237.4,  c  =  1941.6,  B  - 
23°  41' 20";  (16)  6  =  237.41,  c  =  556.82,  ^  =  85°  45' 35";  (17)  ^  =  37°  41', 
B  =  49°  32',  c  =  385.9  ;  (18)  B  =  47°  21'  30",  C  =  81°  49'  45",  6  =  374.26. 

11.  (a)  If  6  :  c  =  11  :  15,  and  A  =  37°  40',  find  B  and  C.  (6)  If  one  side 
of  a  triangle  be  five  times  the  other,  and  their  included  angle  be  64°,  find  the 
remaining  angles. 

12.  (a)  In  ABC,  if  a  :  6  :  c  =  8  :  7  :  5,  find  the  angles.  (6)  The  sides  of 
a  triangle  are  proportional  to  the  numbers  4,  5,  6.     Find  the  least  angle. 

13.  Given  a  =  2  6,  C  =  120°,  find  A,  B,  and  the  ratio  c  :  a. 

14.  (a)  Two  adjacent  sides  of  a  parallelogram  are  respectively  equal  to  12 
and  20  in.,  and  a  diagonal  is  equal  to  25  in.  Find  the  angles  of  the  parallel- 
ogram, the  other  diagonal,  and  the  area.  (6)  The  sides  of  a  quadrilateral 
taken  in  order  are  8,  10,  16,  18,  and  one  diagonal  is  18.  Find  its  angles 
and  area. 

15.  A  ladder  52  ft.  long  is  set  20  ft.  in  front  of  an  inclined  buttress,  and 
reaches  46  ft.  up  its  face.     Find  the  inclination  of  the  face  of  the  buttress. 

16.  A  privateer  is  lying  10  mi.  W.S.  W.  of  a  harbour,  when  a  merchant- 
man leaves  it,  steering  S.E.  8  mi.  an  hour.  If  the  privateer  overtakes  the 
merchantman  in  2  hr.,  find  her  course  and  rate  of  sailing. 

17.  A  fort  bore  E.  by  N.  from  a  beacon,  and  was  distant  from  it  1500  yd. 
From  a  ship  at  anchor  the  beacon  bore  N.N.W.  and  the  fort  N.E.  by  N. 
How  far  was  the  ship  from  the  beacon  ? 


QUESTIONS  AND  EXERCISES,  191 

18.  A  and  B  are  two  points,  200  yd.  apart,  on  the  bank  of  a  river,  and  C 
is  a  point  on  the  opposite  bank ;  the  angles  ABG,  BAG  are  respectively 
54°  30'  and  65°  30'.     Find  the  breadth  of  the  river. 

19.  (a)  Two  observers  on  the  same  side  of  a  balloon,  and  in  the  same 
vertical  plane  with  it,  are  a  mile  apart,  and  they  find  the  angles  of  eleva- 
tion to  be  22°  18'  and  75°  30',  respectively.  What  is  the  height  ?  (6)  Two 
observers  on  opposite  sides  of  a  balloon,  and  in  the  same  vertical  plane  with 
it,  take  its  altitude  simultaneously  ;  one  observer  finds  it  to  be  64°  15',  and 
the  other,  48°  20'.    Find  the  height  of  the  balloon  at  the  time  of  observation. 

20.  From  a  ship  sailing  along  a  coast  a  headland,  C,  was  observed  to 
bear  N.E.  by  N.  After  the  ship  had  sailed  E.  by  N.  15  mi.  the  headland 
bore  W.N.  W.    Find  the  distance  of  the  headland  at  each  observation. 

21.  From  a  certain  station  a  fort,  A,  bore  N.,  and  a  second  fort,  J5,  N.E. 
by  E.  Guns  are  fired  simultaneously  from  the  two  forts,  and  are  heard  at 
the  station  in  1.5  sec.  and  2  sec.  respectively.  Assuming  that  sound  travels 
at  the  rate  of  1142  ft.  per  second,  find  the  distance  of  the  two  forts  apart. 

22.  From  a  point  A  in  the  same  plane  as  the  base  of  a  tower,  the  tower 
bears  N.  62°  W.,  and  the  angle  of  elevation  of  the  top  of  the  tower  is  53°  37' ; 
from  B,  165  ft.  due  north  of  A,  the  tower  bears  west.  What  is  the  height  of 
the  tower  ? 

23.  From  a  ship  steering  W.  by  S.  a  beacon  bore  N.N.W.,  and  after  the 
ship  had  sailed  12  mi.  farther,  the  bearing  of  the  beacon  was  N.E.  by  E. 
At  what  distance  had  the  ship  passed  the  beacon  ? 

24.  From  the  intersection  of  two  straight  paths  which  are  inclined  to  each 
other  at  an  angle  of  37°,  two  pedestrians,  A  and  B,  start  at  the  same  instant 
to  walk  along  the  paths,  A  at  the  rate  of  5  mi.  an  hour,  and  B  at  a  uniform 
rate  also  ;  after  3  hr.  they  are  9^  mi.  apart.  Show  that  there  are  two  rates 
at  which  B  may  walk  to  fulfil  this  condition,  and  find  both  of  those  rates. 

25.  Two  straight  railroads  are  inclined  to  each  other  at  an  angle  of  22°  15'. 
At  the  same  instant  two  engines,  A  and  B,  start  from  a  station  at  the  point 
of  intersection,  A  going  on  one  road  at  the  rate  of  20  mi.  an  hour,  and  B 
going  uniformly  on  the  other.  After  3  hr.  A  and  B  are  25  mi.  apart.  Show 
that  there  are  two  rates  at  which  B  may  go  to  fulfil  this  condition,  and  find 
those  rates. 

26.  A  tower  stood  at  the  foot  of  an  inclined  plane  whose  inclination  to  the 
horizon  was  9°  ;  a  line  was  measured  straight  up  the  incline  from  the  foot  of 
the  tower  of  100  ft.  in  length,  and  at  the  upper  extremity  of  this  line  the 
tower  subtended  an  angle  of  54°.     Find  the  height  of  the  tower. 

27.  The  altitude  of  a  certain  rock  is  observed  to  be  47°,  and  after  walking 
toward  it  1000  ft.  up  a  slope  inclined  at  32°  to  the  horizon,  the  observer  finds 
that  this  altitude  is  77°.  Find  the  vertical  height  of  the  rock  above  the  first 
point  of  observation. 


/ 


V 


102  PLANK   TRIGONOMETRY, 

28.  If,  from  a  point  at  which  the  elevation  of  the  observatory  on  Ben 
Nevi8  Ib  60°,  a  man  walks  800  ft.  on  a  level  ijlane  toward  the  mountain,  and 
then  800  ft.  furthcir  up  a  Klope  of  30"  to  a  point  at  which  the  elevation  of  the 
observatory  is  75°,  show  that  the  height  of  Ben  Nevis  is  approximately 
4478  ft.,  the  man's  path  being  always  supposed  to  lie  in  a  vertical  plane 
passing  through  the  observatory. 

29.  A  man  walks  40  ft.  in  going  straight  down  the  slope  of  the  embank- 
ment of  a  railway  which  runs  due  east  and  west,  and  then  walks  20  ft.  along 
the  foot  of  the  embankment ;  he  finds  that  he  is  exactly  N.E.  of  the  point 
from  which  he  started  at  the  top  of  the  bank.  Show  that  the  inclination  of 
the  bank  to  the  horizon  is  Q0°. 

30.  A  man  in  a  ship  at  sea  sailing  north  observes  two  rocks,  A  and  B,  to 
bear  25°  east  of  his  course  ;  he  then  sails  in  a  direction  northwest  for  4  mi., 
and  observes  A  to  bear  east  and  B  northeast  of  his  new  position.  Find  the 
distance  from  A  to  B. 

31.  To  determine  the  distance  of  two  forts,  C,  Z),  at  the  mouth  of  a  harbour, 
a  boat  is  placed  at  A,  with  its  bow  toward  a  distant  object  E,  and  the  angles 
CAD,  DAE  are  observed  and  found  to  be  22°  17'  and  48°  1'  respectively. 
The  boat  is  then  rowed  to  B,  a  distance  of  1000  yd.,  directly  toward  JE",  and 
the  angles  CBD,  DBE  are  observed  to  be  53°  16'  and  75°  43'  respectively. 
Find  the  distance  CD. 

32.  A  fort  stands  on  a  horizontal  plane  ;  the  angle  of  depression  meas- 
ured from  the  top  of  the  fort  to  a  point  i*on  the  plane  is  ^°,  and  to  a  point  B, 
a  feet  beyond  P,  is  BF'.  Derive  the  formulas  for  computing  h,  the  height  of 
the  fort,  and  rf,  the  distance  from  P  to  the  bottom  of  the  fort. 

83.  A  person  standing  on  a  level  plain  at  the  base  of  a  hill  wishes  to  find 
the  height  of  a  tower  which  is  in  full  sight  on  the  top  of  the  hill.  Describe 
in  detail  the  necessary  measurements  and  computations. 

34.  A  surveyor  wishes  to  find  the  distance  and  the  height  of  a  tower 
which  is  on  the  same  level  with  him,  but  on  the  opposite  side  of  an  impass- 
able chasm ;  illustrate  the  problem  by  a  lettered  figure,  and  describe  in  de- 
tail the  necessary  measurements  and  computations. 

35.  What  measurements  must  be  made  by  an  observer  on  the  shore  to 
find  the  distance  between  two  buoys  ?    Give  formulas  necessary  for  solving. 

36.  Explain  what  measurements  have  to  be  made  at  two  stations,  A  and 
i?,  in  order  to  find  the  distance,  CD,  between  two  inaccessible  objects  {A,  B, 
C,  D  being  in  one  plane) ;  and  state  clearly  the  steps  of  the  calculation  by 
which  the  distance  is  to  be  found  therefrom, 

37.  (a)  From  the  law  of  sines  deduce  that  6  cos  O  +  c  cos  ^  =  a. 
(6)  Prove  this  geometrically,     (c)  Show  that  a  cos  i;  -  &  cos  ^  = . 


QUESTIONS  AND  EXERCISES.  193 

88.   In  any  triangle  ABC,  prove  :  (a)  tan  B  =     ^  ^^^  ^     ; 

a  —  h  cos  C 

X         gg  +  ^2  -  q5  COS  O        _      a     ,  .V  1  4-  cos(^  -  /i)cos  0_  a^+  6^, 

asin  ^  +  6  sin  J5  +  c  sin  C     2  «in^  *         1  +  co8(^  -  6')co8  B     a^  +  c^' 


CHAPTER  VIII. 

[In  what  follows,  8  denotes  the  area  of  a  triangle,  «  its  semi-perimeter, 
R,  r,  Ta,  Th,  n,  the  radii  of  its  circumscribing,  inscribed,  and  escribed  circles, 
respectively.] 

1.  Prove  that  any  side  of  a  triangle  is  equal  to  the  second  side  into  the 
f:fjsine  of  the  angle  opposite  the  third  sine  plus  the  third  side  into  the  cosine 
of  the  angle  opposite  the  second  side. 

2.  Derive  expressions,  in  terms  of  the  sideg  of  a  given  triangle,  for  the 
rjidii  of  its  circumscribing  circle,  and  of  the  four  circles  which  touch  the  side*. 

8.  Derive  expressions  for  the  radii  in  Ex.  2,  in  terms  of  the  sides  and  the 
area  of  the  given  triangle. 

4L  Prove  B  =  ^,  r  =  -,  ra  =  — —    Write  similar  expressions  for  n,  r« 
^S  8  8  —  a 


6.   Prove :  (a)  Va  +  n  -\- Ve  —  r  =  iR  ;  (6)  Vr  •  r«  •  n  •  >*«  =  S. 
6.   Prove:  (a)l  +  i+i  =  l;     (6)  Rr  =  ^^^ 


ra     n     Tc     r'    "•  '  4(a+6  +  c)' 

(c)r  =  458in:^sin^8in£ 
^  ^  2        2        2 

7.  Prove  r«  cot  —  =  n  cot  —  =  r«,  cot  ^  =  r  cot  —  cot  ~  cot  — 

2  2  2  2         2         2 

Ann 

=  4  ^  cos  -  cos  —  cos  -• 
2         2         2 

8.  Prove  B  =  — - — ,  r  =  («  -  a)  tan  — ,  r«  =  « tan  — •    Write  two  other 

2sinvl  ^  ^         2  2 

similar  formulas  for  B  and  r.     Write  similar  formulas  for  r*,  Ve. 

a  sin— sin  — 
2         2 

9.  (a)  Prove  r= ^ •    Write  two  similar  formulas  involving 

cos-  6,  c. 

a  cos— cos— 
2         2 
(h)  Prove  r^  = ^ Write  similar  formulas  for  n,  r*.. 

cos  — 
2 

10.  Show  that  the  length  of  the  tangents  to  the  inscribed  circle  from  the 
angle  ^  is  «  —  a,  from  the  angle  J?  is  «  —  6,  from  the  angle  0  is  «  —  c. 

11.  Write  and  derive  the  formula  for  the  area  of  a  triangle  :  (a)  in  terms 
of  the  three  sides ;  (6)  in  terms  of  two  sides  and  their  included  angle ;  (c)  in 
terms  of  one  side  and  the  two  angles  a^ijacent  to  it. 


194  PLANE  TRIGONOMETRY, 

12.  (a)  Prove  that  /S'  =  s  (s  -  c)  when  O  =  90° ;  (&)  if  x,  y,  be  the  lengths 
of  the  two  diagonals  of  a  parallelogram,  and  6  the  angle  between  them,  show 
that  area  =  \xy  sin 6. 

13.  Find  the  areas  of  some  of  the  triangles  in  Ex.  10,  Chap.  VII.  Find 
the  radii  of  their  circumscribing,  inscribed,  and  escribed  circles. 

14.  (a)  An  isosceles  triangle  whose  vertical  angle  is  78°  contains  400 
square  yards ;  find  the  lengths  of  the  sides.  (6)  Find  two  triangles  each  of 
which  has  sides  63  and  55  ft.  long,  and  an  area  of  874  sq.  ft.  (c)  The  angles 
at  the  base  of  a  triangle  are  22°  30'  and  112°  30'  respectively  ;  show  that  the 
area  of  the  triangle  is  equal  to  the  square  of  half  the  base. 

15.  (a)  Show  that  the  area  of  a  regular  polygon  inscribed  in  a  circle  is  a 
mean  proportional  between  the  areas  of  an  inscribed  and  circumscribing 
polygon  of  half  the  number  of  sides.  (&)  The  sides  of  a  triangle  are  as 
2:3:4;  show  that  the  radii  of  the  escribed  circles  are  as  ^  :  | :  1. 

16.  Two  roads  form  an  angle  of  27°  10'  25".  At  what  distance  from  their 
intersection  must  a  fence  at  right  angles  to  one  of  them  be  placed  so  as  to 
enclose  an  acre  of  land  ? 

17.  If  the  altitude  of  an  isosceles  triangle  is  equal  to  its  base,  the  radius 
of  the  circumscribing  circle  is  f  of  the  base. 

18.  An  equilateral  triangle  and  a  regular  hexagon  have  the  same  perim- 
eter.    Show  that  the  areas  of  their  inscribed  circles  are  as  4  :  9. 

19.  If  the  sides  of  a  triangle  are  51,  68,  and  85  ft.,  show  that  the  shortest 
side  is  divided  by  the  point  of  contact  of  the  inscribed  circle  into  two  seg- 
ments, one  of  which  is  double  the  other. 

CHAPTER  IX. 

1.  Explain  how  angles  are  measured  (1)  by  sexagesimal  measure,  (2)  by 
radian  measure.  Show  how  to  connect  the  radian  measure  of  an  angle  with 
its  measure  in  degrees.  Find  the  number  of  degrees  in  the  angle  called  the 
radian.  How  many  degrees  are  there  in  an  arc  whose  length  is  equal  to 
the  diameter  ?  Show  that  the  radian  measure  of  an  angle  is  the  ratio  of  the 
lengths  of  two  lines.     What  advantage  is  there  in  using  radian  measure  ? 

2.  (a)  Give  the  number  of  degrees  in  each  of  the  following  angles  :  ^  tt, 

ItT,    27r,    T^^TT,    WTT,    1,    -|7r,    fTT,    |  TT,    f  TT,    |  TT,    |  TT,   f  TT,    -|,    -  |  TT,    -  JjW, 

(D^,  (2|)('-),  (—  !)(»•).  (6)  Give  the  supplements  and  complements  of  those 
angles  in  radian  measure  and  in  degree  measure.  (c)  Give  the  radian 
measures  of  30°,  80°,  49°,  41°  30'  15",  120°,  -  210°,  -  175°.  Give  the  radian 
measures  of  their  supplements  and  complements. 

3.  (a)  A  central  angle  1.25'-  is  subtended  by  a  circular  arc  of  16  ft.  ; 
find  the  radius.  (&)  Find  the  number  of  radians  and  degrees  in  the  central 
angle  subtended  by  an  arc  9  in.  long,  iu  a  circle  whose  radius  is  10  ft. 


QUESTIONS  AND  EXERCISES.  195 

^c)  Find  the  radius  of  a  circle  in  which  an  arc  15  in.  long  subtends  at  the 
centre  an  angle  containing  71°  36'  3".6.  (d)  If  the  radius  be  8  in.,  find  the 
central  angle,  in  degrees  and  in  radians,  that  is  subtended  by  an  arc  15  in. 
long,  (e)  An  angle  of  S*"  is  subtended  by  an  arc  of  5  in.  ;  find  the  length  of 
the  radius ;  find  also  the  number  of  radians,  and  of  degrees,  in  an  arc  of 
1.5  in.  (/)  Find  the  number  of  radians  and  seconds  in  the  angle  subtended 
at  the  centre  of  a  circle  whose  radius  is  2  mi.,  by  an  arc  11  in.  long,  (g)  Find 
the  length  of  the  arc  which  subtends  a  central  angle  of  (1)  2  radians,  the 
radius  being  10  in.  ;  (2)  1.5  radians,  radius  2  ft.  ;  (3)  4.3  radians,  radius  21 
yd.  ;  (4)  1.25  radians,  radius  8  in. 

4.  The  value  of  the  division  on  the  outer  rim  of  a  graduated  circle  is  5', 
and  the  distance  between  the  two  successive  divisions  is  .  1  of  an  inch.  Find 
the  radius  of  the  circle. 

5.  Show  that  the  distance  in  miles  between  two  places  on  the  equator, 
which  differ  in  longitude  by  3°  9',  assuming  the  earth's  equatorial  radius  to 
be  7925.6  mi.,  is  217.954  mi. 

6.  (a)  The  difference  of  two  angles  is  10'^,  the  radian  measure  of  their 
sum  is  2.  Find  the  radian  measure  of  each  angle.  (&)  One  angle  of  a  tri- 
angle is  TT  degrees,  another  is  tt  grades.    Show  that  the  radian  measure  of  the 

third  angle  is  tt  —   ^  •     (c)  If  the  number  of  degrees  in  an  angle  be  equal 

to  the  number  of  grades  in  the  complement  of  the  same  angle,  prove  that  the 

radian  measure  of  the  angle  is  — •     (d)  The  angles  of  a  triangle  are  in  the 

ratios  1:2:3.  Express  their  magnitudes  in  each  of  the  three  systems  of 
angular  measurement,  (e)  One  angle  of  a  triangle  is  45°,  another  is  1.5 
radians.  Find  the  third,  both  in  degrees  and  in  radians.  (/)  Express  in 
degrees  and  in  radian  measure  the  vertical  angle  of  an  isosceles  triangle  which 
is  half  of  each  of  the  angles  at  the  base. 

7.  Prove  the  following  statements,  in  which  a  denotes  the  length  of  a 
side  of  a  regular  polygon  ;  P,  the  length  of  its  perimeter  ;  n,  the  number  of 
its  sides  ;  r,  the  radius  of  the  inscribed  circle  ;  i?,  the  radius  of  the  circum- 
scribing circle  : 

a  =  2i?sin-  =  2rtan-;   P=  2ni?sin -  =  2wrtan  -  ;    P  +  r  =  -cot  — ; 
n  n  n  n  2       2n 

area  of  polygon  =Vl B"^  sm  — z=z  nr^ tan  -  =  —  cot -. 
2  n  n       4         n 


CHAPTER   X. 

1.  Define  and  illustrate  the  trigonometric  functions.     Show  in  tabular 
form  the  signs  of  these  functions  in  each  of  the  four  quadrants. 

2.  (a)  Construct  a  table  showing  the  values,  with  proper  signs,  of  the 
trigonometric  functions  of  0°,  30°,  45°,  60°,  120°,  180°,  225°,  270°,  315°,  360°. 


196  PLANE  TRIGONOMETRY, 

(6)  Compare  the  trigonometric  functions  of  90°  -  A,  90°  +  A,  180°  +  A, 
180°  -A,  -A,  with  those  of  A. 

3.  Show,  from  both  the  ratio  and  the  line  definitions  of  the  trigonometric 
functions,  that  (1)  the  sine  and  cosine  are  never  greater  than  unity,  (2)  the 
cosecant  and  secant  are  never  less  than  unity,  (3)  the  tangent  and  cotangent 
may  have  any  values  whatever  from  negative  infinity  to  positive  infinity, 
(4)  the  trigonometric  functions  change  signs  in  passing  through  zero  or 
infinity,  and  through  no  other  values. 

4.  Given  tan  J.  =  —  ^-^,  find  the  values  of  the  other  trigonometric  func- 
tions of  A. 

5.  Find  geometrically  an  expression  for  the  cosine  of  the  difference  of  two 
angles  in  terms  of  the  trigonometric  functions  of  those  angles. 

6.  Prove  that : 

(a)  sin2  B  +  sin2  (A-  B)  +  2  sin  B  sin  (A  -  B)  cos  A  =  sin2  A ; 

cos  nA  -  cos  (n  + 2)  A  ^  ^^^  ^^      ^^^^ 
^  ^  sin  (w  +  2)  A  -  sin  nA  ^  ^ 

7.  Give  the  ratio  definitions  of  the  trigonometric  functions,  sine,  cosine, 
tangent,  and  secant.  These  functions  have  also  been  defined  as  straight 
lines.  Give  these  definitions,  and  show  from  them  that  tan  90°,  sec  90°  would 
each  be  infinite.     Show  that  the  two  systems  are  consistent. 

8.  (a)  Trace  the  changes,  in  magnitude  and  sign,  in  the  values  of  the 
trigonometric  functions  as  the  angle  increases  from  0°  to  360°.  (6)  Trace 
the  changes  of  sign  of  sin  0  as  ^  increases  through  360°,  and  show  that  its 

6          6 
equivalent  2  sin  -  cos  -  has  always  the  same  sign  as  sin  d. 

9.  Trace  the  changes,  as  A  increases  from  0°  to  180°,  in  the  sign  and 
value  of  (a)  cos  (tt  sin^),  (6)  sin  J.  +  cos^,  (c)  sin^  —  cos  A  Draw  the 
graphs  of  these  functions. 

10.  Show  that  the  radian  measure  of  an  acute  angle  is  intermediate  in 
value  between  the  sine  and  the  tangent  of  the  angle. 

11.  (a)  Show  that  the  limit  of  ^^—.  when  B  is  indefinitely  diminished  is  1, 

i.e.  sin  ^  =  ^,  very  nearly.     (6)  Show  that  the  limit  of  ^^^  when  e  is  indefi- 
nitely diminished  is  1,  i.e.  tan  ^  =  0,  very  nearly. 

12.  Find  the  area  of  a  regular  polygon  of  n  sides  inscribed  in  a  circle, 
and  show,  by  increasing  the  number  of  sides  of  the  polygon  without  limit, 
how  the  expression  for  the  area  of  the  circle  may  be  obtained. 

13.  (a)  Find  the  distance  at  which  a  building  50  ft.  wide  will  subtend  an 
angle  of  8'.  (6)  A  church  spire  45  ft.  high  subtends  an  angle  of  9'  at  the 
eye.  Find  its  distance  approximately,  (c)  Find  approximately  the  distance 
of  a  tower  51  ft.  high  which  subtends  at  the  eye  an  angle  of  5y\'.  (d)  How 
large  a  mark  on  a  target  1000  yd.  off  will  subtend  an  angle  of  1"  at  the 
eye? 


QUESTIONS  AND  EXEBCISES,  197 

14.  Show  how  the  functions  may  be  represented  by  lines  connected  with 
a  circle. 

15.  Explain,  with  illustrations,  how  functions  may  be  graphically  repre- 
sented by  means  of  a  curve.    Draw  the  graphs  of  the  trigonometric  functions. 

Note, — "If  a  function  of  a  variable  has  its  magnitude  unaltered  when 
the  sign  of  the  variable  is  changed,  that  function  is  called  an  even  function^ 
but  if  the  function  has  the  same  numerical  value  as  before,  but  with  opposite 
sign,  then  that  function  is  called  an  odd  function  ;  for  instance,  x^  is  an  even 
function  of  ic,  x^  is  an  odd  function  of  x,  but  x^  +  x^  is  neither  even  nor  odd, 
since  its  numerical  value  changes  when  the  sign  of  x  is  changed." 

16.  Show  that  the  cosine,  secant,  and  versine  of  an  angle  are  even  func- 
tions, and  the  sine,  tangent,  cotangent,  and  cosecant  are  odd  functions,  and 
the  coversine  is  neither  even  nor  odd.    (See  Art.  78.) 

CHAPTER  XI. 

N.B.  The  problems  which  are  purely  numerical  are  to  be  solved  inde- 
pendently of  tables.     The  results  can  be  verified  by  means  of  the  tables. 

1.  (a)  Deduce  a  general  expression  for  all  angles  which  have  the  same 
sine  ;  {h)  for  all  which  have  the  same  cosine  ;  (c)  for  all  which  have  the  same 
tangent,  (d)  What  are  the  general  expressions  for  all  angles  which  have  the 
same  secant,  cosecant,  cotangent,  respectively  ? 

2.  Define  inverse  trigonometric  functions ;  give  illustrations.  Define 
tan-ix,  cos-ix. 

3.  (a)  Explain  fully  the  equations  sin(sin-i  1)=^,  sin-i(sin  0')  =  0.  (&)  Con- 
struct sin-i  (f),  cos-i  0,  tan-i  oo,  sec-i^sec  ~\ .     (c)  Find  tan  (cos-i  |). 

[Carefully  state  the  limitations  under  whMsh  the  following  equations  are 
true.  ] 

4.  Show  that :  (a)  tan-i  x+ tan-i  y =tan-i  ^  +  ^  ;   (&)  tan-i  x  -  tan'i  y= 

_  1  -xy 

tan~i  — — ^ ;  (c)  sin-i  x  —  sin-i  y  =  cos-i  Vl  —  x^  —  y^  +  x%2  +  xy. 
I  -\-xy 

6.  (a)  From  sin  2^=2  sin ^  cos ^,  show  that  2  sin-i  x=sin-i(2  xVl  -x^). 
(&)  Show  that  for  certain  values  of  the  angles,  2  cos~ix  =  cos~i(2x2  —  1); 

2  X  ^    .     o 

2  tan-i  X  =  tan-i — ;  2  cot-i  x  =  cosec" 


1  -  x-^  2  X 

:  (a')  cos-ix=sin-H/- 


6.    Show  that  for  certain  values  of  the  angles  :  (a)  cos"i  x=sin-i'v/ + 

cos-iJ^-i-^;     (6)  sin-i J— ^—  =  tan"! V^  =  -  cos-i  ^^—^.      (c)   tan-im-v 
^     2     '     ^  ^  ^a  +  x  ^a     2  a  +  x       ^  ^ 


cot-im=^,or  ^ 
2'         2 


198  PLANE  TRIGONOMETRY. 

7.  Prove  that  for  certain  values:  (1)  sec-i3=2cot-iV2;  (2)  sec2(tan-l2)  + 
cosec2(cot-l3)=15  ;    (3)  sin-if+sin-if=z90°;   (4)  cos-iVj-cos-i  ^^"^-^^ 

2v^3 
^;  (5)  cos-iff+2tan-i^=sin-if;  (6)  cos-ii+2  sin-i|=:120^;  (7)  tan-i^+ 

cot-i  1  +  sin-i  ^  =  0  ;  (8)  tan-i  ^  +  ^  +  tan-K/^  =  ^• 

10  V3-\/2  ^2       4 

8.  Prove  that :  (1)  sin-i    ^^^    +  sin-i  ^^  ~  ^^^  =  ^ ;    (2)  tan-i  (cot  A)  - 


to2  +  71^  m^  +  n^      2 


2l__t...-ill 


tan-i (tan A)  =nT  +--2A;        (3)    tan-i « +  tan-i -^~  =  tan  , 

(4)  tan-i  ^^^Lzil  +  tan-i ^ =  wtt  +  Z^ ;  (5)  tan-i  ^iLl^  +  tan-i  ^^^H^  z:z 

m  2m- 1  4  &^3  ay/3 

- ;  (6)  tan-i  ^  +  tan-i  n  =  cos"! 'i^  -  mn 


3  V(H-m-0(l  +  n'-2) 

9.  Prove  that:  (1)  tan-i -^  +  tan-i -^  +  tan-i ^  =  wtt  ;  (2)  tan-il  = 

1  +  a  1  —  a  or 

tan-i  i  +  tan-i  \  =  tan-i  ^  _j_  tan-i  ^  +  tan"!  i  ;  (3)  tan-i  I  +  tan-i  ^+tan-i  |-+ 

tan-i  I  =  45° ;  (4)  cofi  1  =  cot-i  3  +  cofi  f ;   (5)  cot-il^^  -  cofii^  = 

1  +  a  I  +  b 

sin-i  ^  ~  ^  ;  (6)  4(cot-ii  +  cosec-iV5)  =  7r;  (7)  a  cos  f  sin-i^^  = 

y/a^  -  &2 ;         (8)    sin-i[^~^  +  ^^^  =  -cos-i^^^;         (9)    sincot-ia  = 
tan  cos-1a/^^-±^  ;  (10)  {tan  (sin-i  a)  +  cot  (cos-i  a)}2  =  2  a  tan  (2  tan-i  a). 

10.  Find  all  the  angles  (i.e.  find  the  general  values  of  the  angles)  which 
satisfy    the    following    equations  :     (1)    sec^  A  =  ^;     (2)    2  tan^  6  =  sec^  6 ; 

(3)tan2<9-sec^=l;  (4)  VS  tan2^+l  =  (l+ V3)tan0  ;  (5)  cos0-sin^=— ; 

\/2 
(6)  2  sinx+2  cosecaj=5  ;    (7)  2sin2y=3tan?/ ;    (8)  cos^+tan  jB=sec  jB  ; 
(9)  3  cos2  ic  +  2  V3  cos  X  =  5.25  ;    (10)  tan  ^  -  2  sin  0  =  0  ;    (11)  4  cot  2  ^  = 
cot2  e  -  tan2  6  ;       (12)    _cosec  C  +  cot  0  =  V3  ;       (13)     cot  ^  -  tan  ^  =  2  ; 
(14)  cosec  ic  =  cot  cc  +  V3. 

11.  Solve  cos  2  x  =  sin  x. 

fSoLUTiON  :    cos2a;  =  cosf— —  X ];  or,  sinf  — —  2a;  1=  sinx.    .•.  ^— x  = 
2w7r  ±2x,  or  --2x  =  WTT  +(-  l)'*x. 

12.  Solve  sin  5  u4  =  sin  11  A 

rSOLUTIONJ    ll^  =  W7r+(-l)«5A      /.  ^=0,  ^,   J,    ....1 
L  lb    o  J 

13.  Find  the  general  solutions  of :    (1)  sin  0  +  cos  0  =  \/2  ;    (2)  sin  4  0  = 
ein  ^  J    (3)  2  C09  3  ^  -  2  sin  0  -  X  =  0  ;   (4)  tan2  ^4-3  cot2  e  =  i;  (5)  sin*  x  - 

_..._.___ ___ __._ _..._.._._.__ __._..__.._ __.__ J 


QUESTIONS  AND  EXERCISES.  199 

cos*x  =  l;  (6)  sin"^2ic— sin2aj=.25;  (7)  tan  J5+cot5=2,  cos?/+cos2  2/+ 
cos3y  =  0  (Suggestion:  cos  t/ +  cos  3  y  =  2  cos  y  cos  2  ?/)  ;  (8)sin^x  = 
cosec  X  —  cot  X ;  (9)  cos  x  +  cos  7  x  =  cos  4  a; ;  (10)  cos  ^  +  cos  ^  J.  = 
cos  I  ^  ;  (11)  cosec  z  =  2sia.z  ;  (12)  2  tan-i  cos  A  =  tan-i  2  cosec  A ; 
(13)  tan  (A  -  15°)  =  |  tan  {A  +  16°) ;   (14)  tan(45°  +  -B)  =  1  +  tan  B. 

14.  Find   x   in  the  following  equations:     (1)    cos-i x  +  cos~i (1  —  x)  = 

cos-i  (  -  X) ;       (2)    tan-i  x  +  tan-i  2  x  =  tan-i  ^^ ;       (3)    tan-i  (x  +  1)  + 

6 

tan-i  (X  -  1)  =  tan-i  ^ ;  (4)     tan-i  ^-iti  +  tan'i  ^-Ili  =  tan-i  ( -  7) ; 

(5)  tan-i ^^^  +  tan-i ^^"^  =  tan-i  — • 
^  ^  X  +  1  2  X  +  1  36 

15.  A  flagstaff  a  feet  high  is  on  a  tower  3  a  feet  high ;  prove  that,  if  the 
observer's  eye  is  on  a  level  with  the  top  of  the  staff,  and  the  staff  and  tower 
subtend  equal  angles,  the  observer  is  at  a  distance  a  V2  from  the  top  of  the 
flagstaff. 

16.  In  any  triangle  ABC,  if  tan  -  =  -,  and  tan  -  =  — ,  find  tan  C  with- 

2      6  2      37 

out  tables.  Verify  the  result  by  means  of  the  tables.  Show  that  in  such  a 
triangle,  a  +  c  =  2  6. 

CHAPTER  XII. 

1.  Explain  the  advantages  of  measuring  angles  by  the  sexagesimal,  cen- 
tesimal, and  radian  methods,  respectively. 

2.  Show  that,  if  x  be  the  radian  measure  of  a  positive  angle  less  than  -, 

then  (a)  cos  x  is  less  than  1  but  greater  than  1  —  ^  x^  ;  (&)  sin  x  is  less  than  x 
but  greater  than  x  —  ^  x^.  By  means  of  (b)  show  how  the  sine  of  10"  may 
be  calculated  approximately. 

3.  (a)  Express  in  terms  of  functions  of  A,  each  of  the  following :  sin  2  A, 
cos  2  A  (three  different  forms) ,  tan  2  A.  (b)  Find  cos  3  A  in  terms  of  cos  A. 
(c)  Find  sin  3  ^  in  terms  of  sin  A.  (d)  Find  tan  3  ^  in  terms  of  tan  A,  and 
from  the  formula  determine  the  numerical  value  of  tan  A  if  3  ^  =  90°. 
(e)  Investigate  a  formula  for  expressing  the  cosine  of  half  an  angle  in  terms 
of  the  sine  of  the  whole  angle  ;  and  if  the  angle  lies  between  270°  and  360°, 
show  which  signs  of  the  roots  must  be  taken. 

4.  Show  that  sin  {A  +  B  -  C)  +  ^m{A  +  G  -  B)  -{■  %m{B  +  C  -  A)  - 
sin  (^  +  -B  +  O)  =  4  sin  ^  sin  B  sin  C. 

5.  If  A  +  B-\-C=  180°  (i.e.  if  A,  B,  C  be  the  three  angles  of  a  tri- 
angle), show  that:  (a)  sin^  +  sin  ^  + sinC  =  4  cos  |^  cos^lf  cos  |  O; 
(b)  tan^+tan^+tanC'=tan^tan5tanO;   (c)  cos^+cos  i?+cos  C-1^ 

A        Ti       r  sm^+sm^+smO 

tan  ^  tan  :^  tan  ^. 
2         2         2 


200  PLANE  TRIGONOkETET, 

6.  If  sin  u4  =  f ,  sin  5  =  ^|,  and  sin  C  =  ^5,  where  ^,  B,  C  are  positive 
angles  and  less  than  90°,  find  sin  {A-\-  B  +  C). 

7.  Assuming  the  equation  cos  3  a;  =  4  cos^  a;  —  3  cos  x,  find  sin  18°. 
[Solution  :  64°  +  36°  =  90°.  .-.  cos  54°  =  sin  36° ;  i.e.  cos  3  .  18°  =  sin  2  .  18°. 

Hence,  4  cos^  18°  -  3  cos  18°  =  2  sin  18°  cos  18°.     .-.  4  cos'^  18°  -  3  =  2  sin  18°. 
On  putting  1  —  sin2  18°  for  cos^  18°,  and  solving  for  sin  18°,  there  is  obtained 

the  result,  sin  18°  =  ^"^.3 

8.  Assuming  the  result  in  Ex.  7,  find  the  other  trigonometric  functions 
of  18°  and  the  functions  of  72°. 

9.  Assuming  the  result  in  Ex.  7,  show  that  cos  36°  z=— — ^i^-  =  sin54°. 

4 
Hence,  deduce  the  other  trigonometric  functions  of  36°  and  54°.     Also, 
deduce  the  trigonometric  functions  of  9°  and  81°.     (The  results  in  Exs.  8,  9, 
can  be  verified  by  means  of  the  tables.) 

10.  Prove  the  formulas  : 

sin  (36°  +  A)-  sin  (36°  -A)-  sin  (72°  +  A)  +  sin  (72°  -A)  =  sin  A, 
cos  (36°  +  A.)+  cos(36°  -A)-  cos  (72°  +  A)-  cos  (72°  -A)  =  cos  A, 
^nd  explain  their  use.     (See  Art.  97.) 

11.  (a)  Show  that  sin2  30°  =  sin  18°  sin  54°.  (&)  Solve  x^  cot  108°  = 
128  sin  72°  cos  18°,  without  tables,  (c)  Find  the  trigonometric  functions 
of  48°. 

[Hint  :  48°  =  30°  +  18°.] 

12.  Two  parallel  chords  of  a  circle  lying  on  the  same  side  of  the  centre  of 
a  circle  subtend  angles  of  72°  and  144°  at  the  centre.  Show  that  the  dis- 
tance between  the  chords  is  equal  to  half  the  radius  of  the  circle,  (a)  using 
tables,  (6)  not  using  tables. 

13.  (a)  Solve  :  (i.)  cos  ^  =  0  ;     (ii.)  sin  jc  +  cos  jc  =  1 ; 

(iii.)  tan  y  +  tan  4  2/  +  tan  7  y  =  tan  y  tan  4  y  tan  7  y. 
(b)  If  tan  0  tan  3  ^  =  -  .4,  find  tan  0,  tan  3  6. 


14.    (a)  If  in  triangle  ABC,   A  =  0  B,  show  that  sinJ5  =  -'vl' 


36 -g 

A  2^      6      ' 

(&)  Given  cosJ[  =  .28,  find  tan—.     Explain  the  reason  of  the  ambiguity 

that   presents  itself   in  the  result.       (c)   If  sin^= — ^— -,  find    tan—. 
(d)  (riven  tan lx  =  2  —  v3,  find  sin x. 
15    Prove  the  following : 

(i.)  tan  A  —  tan  IA  =  tan  ^yl  sec  A. 
(ii.)  1  +  tan6  A  =  sec*  A  (sec2  A-S  sin2  A). 

(iii.)  sin  4  +  sin  3  J.  +  sin  5  ^  +  sin  7  yl  =  16  sin  A  cos2  A  cos2  2  A. 
(iv.)  cos  6  4  =  16  (cos6  A  -  sin^  A)  -  15  cos  2  A. 
(v.)  sin  3  ic  +  sin  5 aj  =  8  sin x  cos2  x  cos  2 x. 


QUESTIONS  AND  EXERCISES.  201 

(vi.)  4  cos8  ^  sin  3  ^  +  4  sin^  ^  cos  3  J.  =  3  sin  4  A. 

(vii. )  cos  20°  cos  40°  cos  80°  =  \. 

(viii.)  sin3  A  +  sin^  (120°  +  ^)  +  sin^  (240°  +  ^)  =  -  |  sin  3  A 

(ix.)  1  +  cos 2  (^  -  B)  cos2B  =  cos2  A  +  cos2  (^  -  2 B). 

(x.)  2  cosec  4  ^  +  2  cot  4  ^4  =  cot  w4  —  tan  A. 

16.  Show  that 

cos  (36°  +  A)  cos  (36°  -A)+  cos  (54°  +  A)  cos  (64*'  -  ^)  =  cos  2  -4 ; 
sin  3  ^  =  4  sin  A  sin  (60°  +  A)  sin  (60°  -  -4). 

17.  Show  that 

2  cos  ^  =  ±  Vl  +  sin^  ±  VI  -  sin^,  2sin^  =  ±Vl  +  sinvl  T  VI  -sin4. 

rSuGGESTiON  :    cos2  —  -f  sin^  —  =  1 ;  2  sin  —  cos  —  =  sin  A,  1 
L  2  2         '  2         2  J 

18.  Prove  that  the  following  equations  are  true  for  certain  values  of  the 
angles : 

(i.)  3  sin-i  X  =  sin-i  (3  a;  -  4  x^). 
(ii.)  3  cos-ix  =  cos-i  (4a:3  -  3x). 

(iii.)  tan-i  x  +  tan-i  y  +  tan-i  z  =  tan-i  ^ -^  V  +  ^  -  ^V^ . 

1  —  xy  —  yz  —  zx 

(iv.)  tan-i  X  +  tan-i  y  +  tan-i  '^  -  x  -  y  -  xy  ^  tt^ 

1+x  +  y  -xy     4 
(v.)  Given  tan  «  =  |,  tan  /3  =  |,  tan  7  =  |,  find  tan  (a  +  /3  +  7). 
(vi.)  tan-i ^  +  tan-i  |  =  |  cos-i  f . 

(vii.)  tan-i  -  =  1  tan-i  f  ■=^'\  =  -  tan"!  f^^V 
^      ^  3     2  V    7    y     3  \  117  j 

(viii.)  sin-i^  +  sin-iA  +  sin-ii^  =  ^. 
^       ^  5  13  65     2 

(ix.)  sin-i  If  =  3  sin-i  |. 

(X.)  tan-i  1  +  tan-i  1  =  2:  =  sin-i—  +  cot-i  3. 
^    ^  2^  3     4  V5 

(xi.)  sin-i  -^  +  cos-i  -^^  +  sin-i  1  =  -^ 
^     ^  V73  V146  2      12 

(xii.)  tan-i  I  =  1  tan-i  J/. 

19.  The  hypotenuse  and  shortest  side  of  a  right-angled  triangle  are  5  ft. 
and  3  ft,,  respectively.  Find  the  length  of  the  perpendicular  from  the  right 
angle  upon  the  hypotenuse,  and  show  that  it  is  inclined  at  sin-i  ^\  to  the 
straight  line  drawn  from  the  right  angle  to  the  middle  point  of  the 
hypotenuse. 

20.  If  a  triangle  ABC  is  to  be  solved  from  given  parts  A,  a,  6,  show  that 
the  solution  is  sometimes  ambiguous ;  and  that  in  such  a  case  the  difference 

of  the  two  values  of  C  is  2  cos-i  5il^. 

a 


202  PLANE  TRIGONOMETRY. 

21.   The  tangent  of  an  angle  is  2.4.    Find  the  cosecant  of  the  angle,  the 
cosecant  of  half  the  angle,  and  the  cosecant  of  the  supplement  of  double  the 


22.  The  angle  of  elevation  of  a  tower  at  a  distance  of  20  yd.  from  its  foot 
is  three  times  as  great  as  the  angle  of  elevation  100  yd.  from  the  same  point ; 
show  that  the  height  of  the  tower  is  300  :  \/7  ft. 

23.  DE  is  a  tower  on  a  horizontal  plane.  ABCD  is  a  straight  line  in 
the  plane.  The  tower  subtends  an  angle  ^  at  -4,  2  ^  at  JB,  and  3  0  at  C.  If 
AB  =  50  ft.,  and  BO  =  20  ft.,  find  the  height  of  the  tower  and  the  distance 
CD. 

24.  A  ship  sailing  at  a  uniform  rate  was  observed  to  bear  N.  30°  57'  30"  E. 
After  20  minutes  she  bore  N.  35°  32'  15"  E.,  and  after  10  minutes  more, 
N.  37°  52'  30"  E.    Find  the  direction  in  which  she  was  sailing. 

[Ans.  S.  44°  38'  E.] 

25.  A  spectator  observes  the  explosion  of  a  meteor,  due  south  of  him,  at 
an  elevation  of  28°  45'.  To  another  spectator,  11  mi.  S.S.W.  of  the  former, 
it  appears  at  the  same  instant  to  have  an  altitude  of  42°  15'  30".  Show  that 
there  are  two  possible  heights  above  the  earth's  surface  at  which  it  may  have 
exploded,  and  find  these  heights.  lAns.  4.33  mi.  or  13.21  mi.] 


ANSWERS   TO  THE  EXAMPLES. 


3>0?C 


N.B.  Not  all  of  the  answers  to  the  exercises  are  given.  In  various  ways, 
the  student  should  test,  or  check,  every  result  that  he  obtains  in  working  the 
problems. 

CHAPTER  I. 

Art  2.  1.  logs  27  =  3,  log4  256  =  4,  logu  121  =  2,  ...,  log^p  =  6, 
2.  23  =  8,  54  =  625,  ...,  n«  =  P.  3.  0,  1,  2,  3,  4,  5,  6,  7,  8.  6.  1,  4,  16, 
64,  256,  1024.       7.    0,  1 ;  1,  2  ;  2,  3  ;  ^,  4  ;  3,  4  ;  0,  -1 ;  -1,  -2  ;  -2,  -3. 

Art.  6.  6(a).  1.4007.  6((?).  .09856.  7(6).  7.2767.  8(a).  7.937. 
10.   9.214.     12.    .6443.     13(a).    3.236.     13(c).    1.5563. 

CHAPTER  II. 

Art.  8.  1.  8,  24,  42,  54,  720,  36  a,  12  6,  c.  2.  •..,  a,  |  |-.  3.  ...,  3  a, 
b,±.     4.   12  a,  4  6,  |.     5.   440,  ^j.  ^ 

Art.  10.     5.    1 :  1200,  1  :  253440,  1  :  1920,  ....    7.   2.446  mi. 

Art.  11.  (In  these  answers,  h,  p,  b  represent  hypotenuse,  perpendicular, 
and  base,  respectively.) 

2.   35°,  7i = 34.86,  p =28.56,  -=.574,  -  =  1.743,  ^=.819,  ^=1.22,^=1.428, 
h  b  h  p  b 

-  =  .7.     3.   65°,  b  =  27.19,  p  =  12.68,  -  =  .906,  -  =  2.37,  ^  =  .423,  -  =  1.1, 
p  h  p  h  b 

|=.466,  -=2.14.      4.   56°  19',  33°  41'  (nearly),  ^  =  54.08,  -=.67,  |=1.6, 

^=1.8,  |=.555,   -=1.2,  |=.833.        5.   41°  25',  48°  35'  (nearly),  i)=39.7, 

|=.75,  -  =  1.33,  f=.66,  -=1.51,  -?=.88,  -=1.13.     6.    50°,  p=59.6,  A=77.8, 
h         '  b  h  p  '  b  p 

|  =  .643,  ^=1.56,  ■?=.766,  -=1.31,  |=1.19,  -=.839. 


P 


Art.  14.  11.  (1)  tangent  is  a  :  Vb-^  -  a^  ;  (2)  tangent  is  Vft^  _  ^2  .  ^  ; 
(3)  sine  is  a  :  Va^  +  b^  ;  (4)  sine  is  b  :  Va^  +  52  .  sine  is  y/a'^  -b'^  :a; 
tangent  is  b  :  Va'-^  -  b'^.      12.   41°  24'  35".      13.    19°  28  16". 

203 


204  PLANE  TRIGONOMETRY, 

Art.  15.     1.    2.28025.      2.    2.3333.      3.    5.846.      9.    2.75.      10.    -.708. 

Art.  18.     15.  90°,  36°  52' 12".     16.45°.     17.  45°,  71°  34'.     18.  63°  7' 48". 
19.    30°,  48°  35' 25".      20.    36°  52' 12",  16°  15' 36". 


CHAPTER   III. 

Art.  21.    5.  ^=65°  14',  6=7.834.    6.  ^=50°  12' 24".    9.  ^=30°  12' 12". 
11.    6  =  215.6.      12.   a  =  '312.23. 


CHAPTER  IV. 

Art.  28.      1.   24.948,  12.71.      2.   58.78. 

Art.  29.  4.  398.19  ft.  5.  228.4,258  ft.  6.  63.88  ft.  7.  276.95  ft. 
10.    86.6,  50.      12.   219.45  ft. 

Art.  30.  1.  26.172,  52.345  mi.,  second  ship  bears  E.  19°42'.l  N.  from 
first.      2.    LB  =  14.197  mi. 

Art.  31.      2,  3.    2392.18  sq.  ft.      5.    22.5  sq.  ft.      6.   435.7  sq.  ft. 

Art.  32.  2.  Base  =  187.9  ft.;  height  =  350.63  ft.;  area  =  32,943  sq,  ft. 
3.   Base  =  358.21  ft.;  height  =  161.26  ft.;  area  =  28,881  sq.  ft. 

Art.  33.  1.  14.54  ft.,  16.13  ft.,  48.45  sq.  ft.,  105.2  sq.  ft.  2.  16.516  ft., 
20.415  ft.,  133.94  sq.  ft.,  318.4  sq.  ft. 

Art.  34  b.  2.  10.954  mi.  3.  96  ft.  4.  14.454  mi.  5.  67.08  mi., 
Dip.  =  57' 39".      6.    140.7  mi.,  Dip.  =  2°  V  53". 

Art.  34  c.    4.   2.852  acres.        5.    12  acres  3  roods  6.45  poles. 

CHAPTER  Y. 

Art.  44.  18.  45°,  135°,  -  225°,  -  315°  ;  45°,  135°,  405°,  495°.  19.  60°, 
240°,  -  120°,  -  300°  ;  60°,  240°,  420°,  600°.  20.  135°,  225°,  -  135°,  -  225° ; 
135°,  225°,  495°,  585°.      21.  150°,  330°,  -  30°,  -  210°  ;  150°,  330°,  510°,  690°. 

CHAPTER  VI. 

Art.  46.     8.  cos  (x  +  y)  =  .7874,  sin  (x  +  y)  =  .6164.    (Verify  by  tables.) 

Art.  47.     3.  sin  (x-y)=-  .1582,    cos  {x  -  y)  =  .9874.  (Verify  by 

tables. ) 

Art.  50.  7.  cps6a;  =  cos'-^Sa;  -  sin23x  =  1  -  2sin23x  =  2cos23a;  -  1, 
sin6x=2sin3a:cos3cc.  9.  sin  f  a;=2sin  f  a;cos|  ic,  cos|cc=cos2|a;— sin^fx 
=  1-2  sin-2  |x  =  2  cos'2  |x  -  1.  10.   cos 6 x  =  \/| V 1  +  cos  1 2 x,  sin  6x  = 

V^Vl  —  cosl2x.        12.   sin|x  =  \/}Vl  —  cosf  X,  cos|x  =  V^Vl  +  cosfx. 


ANSWERS.  205 


CHAPTER  VII. 

Art.  55.     2.    6  =  70.8,  a  =  56.1.       4.   6  =  185,  c  =  192.       5.  6  =  8.237, 
;  =  5.464. 

Art.  56.    2.    5  =  36°  18.4'  or  143°  41.6',  c  =  52.71  or  5.98.  b.   A  = 

t8°25'  or  131°  35'. 

Art.  57.    3.   ^  =  80°  46.44',  C  =  63°  48.56'.  4.   i?  =  33°  3.33', 

9  =100°  56.67',  6  =  39.56. 

Art.  58.     2.    16°  47.3',  58°  46.07'.        3.   48°  11.4',  58°  24.7',  73°  23.9'. 

Art.  60.     2.    6  =  698..3,  c  =  845.  3.    600,240.  6.   6  =  749.1. 

t.   B  =  46°  52'  10",    C  =  111°  53'  25",  c  =  1767.3,  or    ^  =  133°  7'  50",    G  = 
J5°37'45",  c  =  823.8. 

Art.  61.     2.   c  =  374.04.        4.   ^  =  109°  15'  30",  c  =  440.46. 

Art.  62.     2.   A  =  53°  7.8',  B  =  59°  29.4'.  4.    P  =  44°  48.25',  B  = 

32°  15.8'. 

Art.  63.     3.   444.72  yd.        4.    1112.8  yd.        6.   179.28  ft.        7.  87.88  ft. 
i.    104.08  ft.        9.   479.8  ft. 


CHAPTER   VIII. 

Art.  67.    1.    11977.8  sq.  ft. ;    46°  13.8',    133°  46.2';     111.3  ft.,   149.1  ft. 
4.   73°  30.7',  106°  29.3' ;  area  =  587637.5  sq.  metres  (approximately). 


CHAPTER  IX; 

_    TT    Sir    lir    5_    11  _    3_    6_        6_    7        3 
11  5 


Art.  73.    2.  (a)  ^,  ^,  ^,  ^,  ^tt,  llir,  ^tt,  -tt,  ..,       „, 

^  ^  4'  3      4  '    6  '   3    '    6     '  2    '   4    '       12    '  20    '  20 


11  TT,   -^TT ;    (6)  .786,  1.048,  2.357,  3.667,  etc.  4.    90°,  60°,  45°,  30°,  36°, 

60  6 

etc.     5.    -135°,  -900°,  -240°,  -165°,  -4500°.  7.   28°  38' 52.4', 

229°  10'  59.2",  171°  53'  14.4",  19°  5'  54.9",  etc.  9.   Sine,  cosine,  tangent, 

cotangent,  secant,  cosecant,  respectively,  are :  -,  -,  — -,  — ^,    "n/3,  — ^,  2  ; 

6    2      2      V3  V3 

f '  v?  ^ '' ''  ^'  ^ ' '' "'  -''  "•  -"'  -''  "'-!"•  f'  I'  ^'  ^s' 

2 
2,  — -.         10  a.    3.75.        11.    The  interior  angles  of  the  polygons  are  respec- 
tively, 3  ^,  I  ^,  5  ^,  3  ^,  4  ^,  I  ^,  ^3  ^.        12.    (1)  20^^  163°  42'  8.3" ;  (2)  ^, 
4°  5'  33.2" ;  (3)  ^,  26".  13.    10,  5,  2.5,  |,  1.25,  |,  |,  i,  20,  30,  25, 

26|  in.        15.    10,  40,  70,  80,  120,  5,  3.75,  1.25  in. 


206 


PLANE  TBIGONOMETRY, 


CHAPTER   X.       , 

Art.  83.     4.    238,890  mi.  (approximately),  347.5.  ,5.    About  57' 2"; 

about  1  :  13.5.         6.    About  93,757,000  mi.         7.    206,265  times  the  distance 
of  the  earth  from  the  sun,  3.26  yr.  8.    9  ft.  2.6  in.  9.    76  ft.  9.5  in. 

10.   4' 35".        11.   15.708  yd.        12.    13' 1.3". 


CHAPTER  XI. 


Art.  85.     3.    ^=W7r+(-l)"-,60°,  120°,  420°,  480°.     4.   n7r+(-l)"+i-, 

3  .  6 

210°,  330°,  570°,  690°.  5.  n  •  180°+  (-1)«  72°  30',  107° 30',  432° 30',  467° 30'. 

Art.  86.  2.  2  WTT  ±  ^,  w  .  360  ±  30°,  30°,  330°-,  390°,  690°.  3.  w  •  360  ± 

7°  40',  7°  40',  352°  20',  367°  40',  712°  20'.    5.  n  -  360°  ±  136°  35',  136°  35', 
223°  25',  496°  .35',  583°  25'. 

Art.  87.  2.  WTT  +  - ,  w  .  180°  +  60°,  60°,  240°,  420°,  600°.  3.  n  •  180°  + 

o 

20°  10',  20°  10',  200°  10',  380°  10',  660°  10'.   6.  n  •  180°  +  138°,  138°,  318°, 

498°,  678°. 

Art.  89.     11  (first  part),  (a)  .97302  ;  (&)  ±.97302,  .±.12117.    12.  43° 6.5'. 
13.   45°. 

Art.  90.     5.    W7r±J,  W7r±J.  6.    W7r  +  -,   W7r  +  f7r  [i.e.  ^  + -V 

D  o  8  \  2        8  / 

7.    w.  180°+ 36° 62.2'.    8.    n  •  180° +  63° 26',  n  •  180°- 71° 33.9'.     9.   ^-180°+ 

63° 26',   (4 n- 1)45°.  10-   ^,   !^  +  (_l)n-E-.  n.    w7r+(-l)n^, 

4  3  18  2 

,1.360° -46° 23.85'.     12.  (2n  +  l)90°,  {4n+(-l)"}15°.    13.  {6n+(-l)«}30°, 

(10w  +  (-l)«}18°,  {10n-3(-l)-}18°.     14.  wtt  ±  ^,  wtt  ±^.    15.  wir±J, 

4  3  6 

W7r±J.      16.    0,  ±\. 

CHAPTER   XII. 

Art    94     9       4  tan  A  —  i  tan^  A         6  tan  ^  —  10  tan^  A  +  tan^  A 
*    1  -  6  tan2  A  +  tan*  A'         1-10  tan2  ^  +  6  tan*  A    ' 
6  tan  ^  -  20  tan^  ^  +  6  tan^  A        7  tan  ^  -  35  tan^  ^  +  21  tan^  A  -  tan'  ^ 
1-15  tan2  A  +  \b  tan*  A  -  tan^  A'       1-21  tan2  ^  +  35  tan*  A-1  tan^  A 


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